2012 ACM/ICPC 亚洲区 金华站

题目链接  2012金华区域赛

Problem A

按a/b从小到大的顺序排队进行体检即可

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstdlib>
 5 #include<algorithm>
 6 #include<cstring>
 7 #include<string>
 8 #include<vector>
 9 #include<map>
10 #include<set>
11 #include<queue>
12 using namespace std;
13 typedef long long ll;
14 const ll mod=365*24*60*60;
15 struct ss
16 {
17     int x,y;
18     double p;
19 };
20 ss a[100010];
21 int n;
22 inline bool cmp(ss a,ss b)
23 {
24     return a.p<b.p;
25 }
26 int main()
27 {
28     while (~scanf("%d",&n))
29     {
30         if (n==0) break;
31         int i;
32         for (i=1;i<=n;i++)
33         {
34             scanf("%d%d",&a[i].x,&a[i].y);
35             a[i].p=a[i].x*1.0/a[i].y;
36         }
37         sort(a+1,a+n+1,cmp);
38         ll sum=0;
39         for (i=1;i<=n;i++)
40             sum=((sum+sum*a[i].y%mod)%mod+a[i].x)%mod;
41         printf("%lld\n",sum);
42     }
43     return 0;
44 }
View Code

 

Problem B

Problem C

先把所有的坐标离散化

将起点、终点和所有矩形的点

每个点4个方向，拆成4个点

然后如果一个点的一个方向能经过一次转弯直接到另一个点

就连一条边

建完图后用SPFA求最短路即可

但本题细节较多

最难处理的是墙角的转角情况

可以先把所有墙角预处理出来

一共8种情况

其他的就离散化后坐标乘2

然后用个数组标记每个点会经过几次

如果点已经在矩形内部

就直接置为经过2次

然后在判断时 判断路线上有没有经过2次以上的点

或者恰有一个墙角点但可以转弯

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cmath>
  4 #include<cstdlib>
  5 #include<algorithm>
  6 #include<cstring>
  7 #include<string>
  8 #include<vector>
  9 #include<map>
 10 #include<set>
 11 #include<queue>
 12 using namespace std;
 13 #define y1 hjhk
 14 #define y2 mhgu
 15 const int inf=(1<<31)-1;
 16 struct Hash : vector<int> {
 17     void prepare() {
 18         sort(begin(), end());
 19         erase(unique(begin(), end()), end());
 20     }
 21     int get(int x) {
 22         return lower_bound(begin(), end(), x)-begin()+1;
 23     }
 24 } has;
 25 int x1,y1,x2,y2,n;
 26 struct ss
 27 {
 28     int x1,y1,x2,y2;
 29 };
 30 ss a[51];
 31 int dis[1500][4];
 32 int can[1500][1500];
 33 bool mop[1500][1500][4][4];
 34 queue<pair<int,int>> q;
 35 bool vis[1500][4];
 36 pair<int,int> det[1500];
 37 void relax(pair<int,int> from,int x2,int y2,int ndis)
 38 {
 39     //cout<<from.first<<" "<<from.second<<" "<<x2<<" "<<y2<<" "<<ndis<<endl;
 40     if (ndis<dis[x2][y2])
 41     {
 42         dis[x2][y2]=ndis;
 43         //cout<<from.first<<" "<<from.second<<" "<<x2<<" "<<y2<<" "<<ndis<<endl;
 44         if (!vis[x2][y2])
 45         {
 46             q.push({x2,y2});
 47             vis[x2][y2]=true;
 48         }
 49     }
 50 }
 51 int main()
 52 {
 53     //freopen("4444.in","r",stdin);
 54     //freopen("out.txt","w",stdout);
 55     while (~scanf("%d%d%d%d",&x1,&y1,&x2,&y2))
 56     {
 57         if (x1==0&&y1==0&&x2==0&&y2==0) return 0;
 58         scanf("%d",&n);
 59         int i,j;
 60         for (i=1;i<=n;i++)
 61         {
 62             scanf("%d%d%d%d",&a[i].x1,&a[i].y1,&a[i].x2,&a[i].y2);
 63             if (a[i].x1>a[i].x2)
 64                 swap(a[i].x1,a[i].x2);
 65             if (a[i].y1>a[i].y2)
 66                 swap(a[i].y1,a[i].y2);
 67         }
 68         has.clear();
 69         has.push_back(x1);
 70         has.push_back(x2);
 71         for (i=1;i<=n;i++)
 72         {
 73             has.push_back(a[i].x1);
 74             has.push_back(a[i].x2);
 75         }
 76         has.prepare();
 77         x1=has.get(x1)*2;
 78         x2=has.get(x2)*2;
 79         for (i=1;i<=n;i++)
 80         {
 81             a[i].x1=has.get(a[i].x1)*2;
 82             a[i].x2=has.get(a[i].x2)*2;
 83         }
 84         has.clear();
 85         has.push_back(y1);
 86         has.push_back(y2);
 87         for (i=1;i<=n;i++)
 88         {
 89             has.push_back(a[i].y1);
 90             has.push_back(a[i].y2);
 91         }
 92         has.prepare();
 93         y1=has.get(y1)*2;
 94         y2=has.get(y2)*2;
 95         for (i=1;i<=n;i++)
 96         {
 97             a[i].y1=has.get(a[i].y1)*2;
 98             a[i].y2=has.get(a[i].y2)*2;
 99         }
100         //cout<<x1<<" "<<y1<<" "<<x2<<" "<<y2<<endl;
101         //for (i=1;i<=n;i++)
102         //    printf("%d %d %d %d\n",a[i].x1,a[i].y1,a[i].x2,a[i].y2);
103         memset(mop,false,sizeof(mop));
104         for (i=1;i<=n;i++)
105             for (j=1;j<=n;j++)
106                 if (i!=j)
107                 {
108                     if (a[i].x2==a[j].x1&&a[i].y1<=a[j].y1)
109                     {
110                         mop[a[j].x1][a[j].y1][1][2]=true;
111                         mop[a[j].x1][a[j].y1][0][3]=true;
112                         //cout<<a[j].x1<<" "<<a[j].y1<<" "<<1<<" "<<2<<endl;
113 
114                     }
115                     if (a[i].y2==a[j].y1&&a[i].x2<=a[j].x2)
116                     {
117                         mop[a[i].x2][a[i].y2][1][2]=true;
118                         mop[a[i].x2][a[i].y2][0][3]=true;
119                         //cout<<a[i].x2<<" "<<a[i].y2<<" "<<1<<" "<<2<<endl;
120                     }
121                     if (a[i].x2==a[j].x1&&a[i].y2<=a[j].y2)
122                     {
123                         mop[a[i].x2][a[i].y2][2][1]=true;
124                         mop[a[i].x2][a[i].y2][3][0]=true;
125                         //cout<<a[i].x2<<" "<<a[i].y2<<" "<<2<<" "<<1<<endl;
126                     }
127                     if (a[i].y2==a[j].y1&&a[i].x1<=a[j].x1)
128                     {
129                         mop[a[j].x1][a[j].y1][2][1]=true;
130                         mop[a[j].x1][a[j].y1][3][0]=true;
131                         //cout<<a[j].x1<<" "<<a[j].y1<<" "<<2<<" "<<1<<endl;
132                     }    
133                     if (a[i].x2==a[j].x1&&a[i].y1>=a[j].y1)
134                     {
135                         mop[a[i].x2][a[i].y1][1][0]=true;
136                         mop[a[i].x2][a[i].y1][2][3]=true;
137                         //cout<<a[i].x2<<" "<<a[i].y1<<" "<<1<<" "<<0<<endl;
138                     }
139                     if (a[i].y2==a[j].y1&&a[i].x1>=a[j].x1)
140                     {
141                         mop[a[i].x1][a[i].y2][1][0]=true;
142                         mop[a[i].x1][a[i].y2][2][3]=true;
143                         //cout<<a[i].x1<<" "<<a[i].y2<<" "<<1<<" "<<0<<endl;
144                     }
145                     if (a[i].y2==a[j].y1&&a[i].x2>=a[j].x2)
146                     {
147                         mop[a[j].x2][a[j].y1][0][1]=true;
148                         mop[a[j].x2][a[j].y1][3][2]=true;
149                         //cout<<a[j].x2<<" "<<a[j].y1<<" "<<0<<" "<<1<<endl;
150                     }
151                     if (a[i].x2==a[j].x2&&a[i].y2>=a[j].y2)
152                     {
153                         mop[a[j].x1][a[j].y2][0][1]=true;
154                         mop[a[j].x1][a[j].y2][3][2]=true;
155                         //cout<<a[j].x1<<" "<<a[j].y2<<" "<<0<<" "<<1<<endl;
156                     }
157                 }
158         memset(can,0,sizeof(can));
159         //cout<<x1<<" "<<y1<<" "<<x2<<" "<<y2<<endl;
160         for (i=1;i<=n;i++)
161         {
162             for (int x=a[i].x1;x<=a[i].x2;x++)
163                 for (int y=a[i].y1;y<=a[i].y2;y++)
164                 {
165                     can[x][y]++;
166                     if (x!=a[i].x1&&x!=a[i].x2&&y!=a[i].y1&&y!=a[i].y2)
167                     {
168                         can[x][y]=2;
169                         //if (x==11&&y==60)
170                         //    cout<<a[i].x1<<" "<<a[i].x2<<" "<<a[i].y1<<" "<<a[i].y2<<endl;
171                     }
172                     if (can[x][y]>2) can[x][y]=2;
173                     //cout<<x<<" "<<y<<" "<<can[x][y]<<endl;
174                 }
175         }
176         for (i=1;i<=n;i++)
177             for (j=1;j<=n;j++)
178                 if (i!=j)
179                 {
180                     if (a[i].x2==a[j].x1&&a[i].y1==a[j].y1) can[a[i].x2][a[i].y1]=1;
181                     if (a[i].x2==a[j].x1&&a[i].y2==a[j].y2) can[a[i].x2][a[i].y2]=1;
182                     if (a[i].y2==a[j].y1&&a[i].x1==a[j].x1) can[a[i].x1][a[i].y2]=1;
183                     if (a[i].y2==a[j].y1&&a[i].x2==a[j].x2) can[a[i].x2][a[i].y2]=1;
184                 }
185         int cnt=0;
186         det[0].first=x1;
187         det[0].second=y1;
188         for (i=1;i<=n;i++)
189         {
190             det[++cnt].first=a[i].x1;
191             det[cnt].second=a[i].y1;
192             det[++cnt].first=a[i].x1;
193             det[cnt].second=a[i].y2;
194             det[++cnt].first=a[i].x2;
195             det[cnt].second=a[i].y1;
196             det[++cnt].first=a[i].x2;
197             det[cnt].second=a[i].y2;
198         }
199         det[++cnt].first=x2;
200         det[cnt].second=y2;
201         for (i=0;i<=cnt;i++)
202             dis[i][0]=dis[i][1]=dis[i][2]=dis[i][3]=inf;
203         dis[0][0]=0;
204         dis[0][1]=0;
205         dis[0][2]=0;
206         dis[0][3]=0;
207         memset(vis,false,sizeof(vis));
208         vis[0][0]=true;
209         vis[0][1]=true;
210         vis[0][2]=true;
211         vis[0][3]=true;
212         while (!q.empty()) q.pop();
213         q.push({0,0});
214         q.push({0,1});
215         q.push({0,2});
216         q.push({0,3});
217         while (!q.empty())
218         {
219             pair<int,int> x=q.front();
220             q.pop();
221             vis[x.first][x.second]=false;
222             //cout<<det[x.first].first<<" "<<det[x.first].second<<endl;
223             //cout<<x.second<<endl;
224             //cout<< dis[x.first][x.second]<<endl;
225             for (i=0;i<=cnt;i++)
226                 if (i!=x.first)
227                 {
228                     int x1=det[x.first].first;
229                     int y1=det[x.first].second;
230                     int x2=det[i].first;
231                     int y2=det[i].second;
232                     //cout<<i<<" "<<x1<<" "<<y1<<" "<<x2<<" "<<y2<<endl;
233                     if (x.second==3)
234                     {
235                         if (x2==x1&&y2<y1)
236                         {
237                             bool fg=true;
238                             for (int y=y2;y<=y1;y++)
239                                 if (can[x1][y]==2)
240                                 {
241                                     fg=false;
242                                     break;
243                                 }
244                             if (fg)
245                                 relax(x,i,3,dis[x.first][x.second]);
246                         }
247                         else if (x2<x1&&y2<y1)
248                         {
249                             int fg=0;
250                             for (int y=y2;y<=y1;y++)
251                                 if (can[x1][y]==2)
252                                 {
253                                     fg++;
254                                 }
255                             for (int x=x2;x<=x1;x++)
256                                 if (can[x][y2]==2)
257                                 {
258                                     fg++;
259                                 }
260                             if (fg==0||(fg==2&&can[x1][y2]==2&&mop[x1][y2][3][0]))
261                                 relax(x,i,0,dis[x.first][x.second]+1);
262                         }
263                         else if (x2>x1&&y2<y1)
264                         {
265                             int fg=0;
266                             for (int y=y2;y<=y1;y++)
267                                 if (can[x1][y]==2)
268                                 {
269                                     fg++;
270                                     //break;
271                                 }
272                             for (int x=x1;x<=x2;x++)
273                                 if (can[x][y2]==2)
274                                 {
275                                     fg++;
276                                     //break;
277                                 }
278                             if (fg==0||(fg==2&&can[x1][y2]==2&&mop[x1][y2][3][2]))
279                                 relax(x,i,2,dis[x.first][x.second]+1);
280                         }
281                         relax(x,x.first,2,dis[x.first][x.second]+1);
282                         relax(x,x.first,0,dis[x.first][x.second]+1);
283                     }
284                     else if (x.second==0)
285                     {
286                         if (x2<x1&&y2==y1)
287                         {
288                             bool fg=true;
289                             for (int x=x2;x<=x1;x++)
290                                 if (can[x][y2]==2)
291                                 {
292                                     fg=false;
293                                     break;
294                                 }
295                             if (fg)
296                                 relax(x,i,0,dis[x.first][x.second]);
297                         }
298                         else if (x2<x1&&y2<y1)
299                         {
300                             int fg=0;
301                             for (int y=y2;y<=y1;y++)
302                                 if (can[x2][y]==2)
303                                 {
304                                     fg++;
305                                     //break;
306                                 }
307                             for (int x=x2;x<=x1;x++)
308                                 if (can[x][y1]==2)
309                                 {
310                                     fg++;
311                                     //break;
312                                 }
313                             if (fg==0||(fg==2&&can[x2][y1]==2&&mop[x2][y1][0][3]))
314                                 relax(x,i,3,dis[x.first][x.second]+1);
315                         }
316                         else if (x2<x1&&y2>y1)
317                         {
318                             int fg=0;
319                             for (int y=y1;y<=y2;y++)
320                                 if (can[x2][y]==2)
321                                 {
322                                     fg++;
323                                     //break;
324                                 }
325                             for (int x=x2;x<=x1;x++)
326                                 if (can[x][y1]==2)
327                                 {
328                                     fg++;
329                                     //break;
330                                 }
331                             if (fg==0||(fg==2&&can[x2][y1]==2&&mop[x2][y1][0][1]))
332                                 relax(x,i,1,dis[x.first][x.second]+1);
333                         }
334                         relax(x,x.first,1,dis[x.first][x.second]+1);
335                         relax(x,x.first,3,dis[x.first][x.second]+1);
336                         
337                     }
338                     else if (x.second==1)
339                     {
340                         if (x2==x1&&y2>y1)
341                         {
342                             bool fg=true;
343                             for (int y=y1;y<=y2;y++)
344                                 if (can[x2][y]==2)
345                                 {
346                                     fg=false;
347                                     break;
348                                 }
349                             if (fg)
350                                 relax(x,i,1,dis[x.first][x.second]);
351                         }
352                         else if (x2<x1&&y2>y1)
353                         {
354                             int fg=0;
355                             for (int y=y1;y<=y2;y++)
356                                 if (can[x1][y]==2)
357                                 {
358                                     fg++;
359                                     //break;
360                                 }
361                             for (int x=x2;x<=x1;x++)
362                                 if (can[x][y2]==2)
363                                 {
364                                     fg++;
365                                     //break;
366                                 }
367                             if (fg==0||(fg==2&&can[x1][y2]==2&&mop[x1][y2][1][0]))
368                                 relax(x,i,0,dis[x.first][x.second]+1);
369                         }
370                         else if (x2>x1&&y2>y1)
371                         {
372                             int fg=0;
373                             for (int y=y1;y<=y2;y++)
374                                 if (can[x1][y]==2)
375                                 {
376                                     fg++;
377                                     //break;
378                                 }
379                             for (int x=x1;x<=x2;x++)
380                                 if (can[x][y2]==2)
381                                 {
382                                     fg++;
383                                     //break;
384                                 }
385                             if (fg==0||(fg==2&&can[x1][y2]==2&&mop[x1][y2][1][2]))
386                                 relax(x,i,2,dis[x.first][x.second]+1);
387                         }
388                         relax(x,x.first,2,dis[x.first][x.second]+1);
389                         relax(x,x.first,0,dis[x.first][x.second]+1);
390                         
391                     }
392                     else if (x.second==2)
393                     {
394                         if (x2>x1&&y2==y1)
395                         {
396                             bool fg=true;
397                             for (int x=x1;x<=x2;x++)
398                                 if (can[x][y2]==2)
399                                 {
400                                     //cout<<x<<" "<<y2<<endl;
401                                     fg=false;
402                                     break;
403                                 }
404                             //cout<<fg<<" "<<cnt<<" "<<i<<endl;
405                             if (fg)
406                                 relax(x,i,2,dis[x.first][x.second]);
407                         }
408                         else if (x2>x1&&y2<y1)
409                         {
410                             int fg=0;
411                             for (int y=y2;y<=y1;y++)
412                                 if (can[x2][y]==2)
413                                 {
414                                     fg++;
415                                     //break;
416                                 }
417                             for (int x=x1;x<=x2;x++)
418                                 if (can[x][y1]==2)
419                                 {
420                                     fg++;
421                                     //break;
422                                 }
423                             if (fg==0||(fg==2&&can[x2][y1]==2&&mop[x2][y1][2][3]))
424                                 relax(x,i,3,dis[x.first][x.second]+1);
425                         }
426                         else if (x2>x1&&y2>y1)
427                         {
428                             int fg=0;
429                             for (int y=y1;y<=y2;y++)
430                                 if (can[x2][y]==2)
431                                 {
432                                     fg++;
433                                     //break;
434                                 }
435                             for (int x=x1;x<=x2;x++)
436                                 if (can[x][y1]==2)
437                                 {
438                                     fg++;
439                                     //break;
440                                 }
441                             if (fg==0||(fg==2&&can[x2][y1]==2&&mop[x2][y1][2][1]))
442                                 relax(x,i,1,dis[x.first][x.second]+1);
443                         }
444                         relax(x,x.first,1,dis[x.first][x.second]+1);
445                         relax(x,x.first,3,dis[x.first][x.second]+1);
446                     }
447                 }    
448         }
449         int ans=inf;
450         ans=min(dis[cnt][0],ans);
451         ans=min(dis[cnt][1],ans);
452         ans=min(dis[cnt][2],ans);
453         ans=min(dis[cnt][3],ans);
454         if (ans==inf) puts("-1");
455         else printf("%d\n",ans);
456     }
457     return 0;
458 }
View Code

 

Problem D

直接把180度分成3000分然后枚举取最小值即可。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)

const double d = 3000;
const double pi = acos(-1.0);
const double g = 9.8000000;
const int N = 2003;


double v[N];
double l1, r1, l2, r2;
int n;
double h;
int ans = 0;
int now;

int solve(double ap){
	int ret = 0;
	rep(i, 1, n){
		double now;
		now = sqrt(v[i] * v[i] * sin(ap) * sin(ap) + 2 * h * g);
		now = (now - v[i] * sin(ap)) / g * v[i] * cos(ap);

		if (now >= l2 && now <= r2) return -1;
		if (now >= l1 && now <= r1) ++ret;
	}

	return ret;
}	

int main(){

	while (~scanf("%d", &n), n){
		ans = 0;
		scanf("%lf%lf%lf%lf%lf", &h, &l1, &r1, &l2, &r2);
		rep(i, 1, n) scanf("%lf", v + i);
		
		for (double i = -0.5 * pi; i <= 0.5 * pi; i += (pi / d)){
			now = solve(i);
			if (~now) ans = max(ans, now);
		}
		

		now = solve(-0.5 * pi);
		if (~now) ans = max(ans, now);
		now = solve(0);
		if (~now) ans = max(ans, now);
		now = solve(0.5 * pi);
		if (~now) ans = max(ans, now);
		printf("%d\n", ans);
	}

	return 0;
}

 

Problem E

Problem F

分解了十几项发现

$f_{x} = g_{a_{1}} * g_{a_{2}} * ... * g_{a_{m}}$，m为x的约数个数。

$a_{1},a_{2},...,a_{m}$分别为x的约数。

我们预处理出$g_{x}，g_{x} = f_{x} / g_{a_{1}} / g_{a_{2}} / ... / g_{a_{m-1}}$

然后就可以计算答案了。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)

const int N = 2200;

vector <int> c[N];

struct dxs{
	int l;
	int d[N];
	dxs(){ memset(d, 0, sizeof d); l = 1;}
}  a[N];

int n;


inline dxs chu(dxs a, dxs b){
	dxs ret;
	ret.l = a.l - b.l + 1;
	memset(ret.d, 0, sizeof ret.d);
	dec(i, a.l - 1, b.l - 1){
		int tt = a.d[i] / b.d[b.l - 1];
		if (tt != 0){
			rep(j, 0, b.l - 1){
				a.d[i - j] -= tt * b.d[b.l - 1 - j];
			}

			ret.d[i - b.l + 1] = tt;
		}
	}
	return ret;

}

bool cmp(int i, int j){
	if (a[i].l != a[j].l) return a[i].l < a[j].l;
	dec(k, a[i].l - 1, 0)
		if(a[i].d[k] != a[j].d[k])
		{
			if(abs(a[i].d[k]) == abs(a[j].d[k]))
				return a[i].d[k] < 0;
			else return abs(a[i].d[k]) < abs(a[j].d[k]);
		}
	return false;
}

void print(dxs a){
	putchar('(');
	dec(i, a.l - 1, 0) if (a.d[i] != 0){
		if (i == 0){
			if (a.d[i] > 0) putchar('+');
			printf("%d", a.d[i]);
		}
		else{
			if (a.d[i] > 0 && i != a.l - 1) putchar('+');
			if (a.d[i] == -1) putchar('-');
			if (abs(a.d[i]) != 1) printf("%d", a.d[i]);
			if (i > 1) printf("x^%d", i);
			else putchar('x');
		}
	}
	putchar(')');
}


int main(){

	rep(i, 2, 1100){
		rep(j, 1, i) if (i % j == 0) c[i].push_back(j);
	}


	a[1].l = 2; a[1].d[0] = -1, a[1].d[1] = 1;
	rep(i, 2, 1100){
		a[i].d[0] = -1; a[i].d[i] = 1; a[i].l = i + 1; a[i] = chu(a[i], a[1]);
		rep(j, 2, i - 1) if (i % j == 0) a[i] = chu(a[i], a[j]);
	}

	rep(i, 2, 1100) sort(c[i].begin(), c[i].end(), cmp);

	while (~scanf("%d", &n), n){
		if (n == 1){
			puts("x-1");
			continue;
		}
		for (auto u : c[n]) print(a[u]);
		putchar(10);
	}


	return 0;
}

 

 

Problem G

Problem H

先求出三维的凸包

然后枚举每个面作为底面

求出旋转后所有点的坐标

最后求出底面二维凸包的面积即可

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cmath>
  4 #include <algorithm>
  5 #define eps 1e-7
  6 using namespace std;
  7 const double inf=0x3f3f3f3f;
  8 const int MAXV=80;
  9 const double EPS = 1e-9;
 10 const double pi=acos(-1.0);
 11 //三维点
 12 struct pt {
 13     double x, y, z;
 14     pt() {}
 15     pt(double _x, double _y, double _z): x(_x), y(_y), z(_z) {}
 16     pt operator - (const pt p1) {
 17         return pt(x - p1.x, y - p1.y, z - p1.z);
 18     }
 19     pt operator + (const pt p1) {
 20         return pt(x + p1.x, y + p1.y, z + p1.z);
 21     }
 22     pt operator *(const pt p){
 23         return pt(y*p.z-z*p.y,z*p.x-x*p.z, x*p.y-y*p.x);
 24     }
 25     pt operator *(double d)
 26     {
 27         return pt(x*d,y*d,z*d);
 28     }
 29     
 30     pt operator / (double d)
 31     {
 32         return pt(x/d,y/d,z/d);
 33     }
 34     double operator ^ (pt p) {
 35         return x*p.x+y*p.y+z*p.z;    //点乘
 36     }
 37     double len(){
 38         return sqrt(x*x+y*y+z*z);
 39     }
 40 };
 41 struct pp{
 42     double x,y;
 43     pp(){}
 44     pp(double x,double y):x(x),y(y){}
 45 };
 46 struct _3DCH {
 47     struct fac {
 48         int a, b, c;    //表示凸包一个面上三个点的编号
 49         bool ok;        //表示该面是否属于最终凸包中的面
 50     };
 51     
 52     int n;    //初始点数
 53     pt P[MAXV];    //初始点
 54     
 55     int cnt;    //凸包表面的三角形数
 56     fac F[MAXV*8]; //凸包表面的三角形
 57     
 58     int to[MAXV][MAXV];
 59     
 60     pt Cross3(pt a,pt p){
 61         return pt(a.y*p.z-a.z*p.y, a.z*p.x-a.x*p.z, a.x*p.y-a.y*p.x);    //叉乘
 62     }
 63     double vlen(pt a) {
 64         return sqrt(a.x*a.x+a.y*a.y+a.z*a.z);    //向量长度
 65     }
 66     double area(pt a, pt b, pt c) {
 67         return vlen(Cross3((b-a),(c-a)));    //三角形面积*2
 68     }
 69     double volume(pt a, pt b, pt c, pt d) {
 70         return Cross3((b-a),(c-a))^(d-a);    //四面体有向体积*6
 71     }
 72     //三维点积
 73     double Dot3( pt u, pt v )
 74     {
 75         return u.x * v.x + u.y * v.y + u.z * v.z;
 76     }
 77     
 78     //平面的法向量
 79     pt pvec(pt a,pt b,pt c)
 80     {
 81         return (Cross3((a-b),(b-c)));
 82     }
 83     //点到面的距离
 84     double Dis(pt a,pt b,pt c,pt d)
 85     {
 86         return fabs(pvec(a,b,c)^(d-a))/vlen(pvec(a,b,c));
 87     }
 88     //正：点在面同向
 89     double ptof(pt &p, fac &f) {
 90         pt m = P[f.b]-P[f.a], n = P[f.c]-P[f.a], t = p-P[f.a];
 91         return Cross3(m , n) ^ t;
 92     }
 93     
 94     void deal(int p, int a, int b) {
 95         int f = to[a][b];
 96         fac add;
 97         if (F[f].ok) {
 98             if (ptof(P[p], F[f]) > eps)
 99                 dfs(p, f);
100             else {
101                 add.a = b, add.b = a, add.c = p, add.ok = 1;
102                 to[p][b] = to[a][p] = to[b][a] = cnt;
103                 F[cnt++] = add;
104             }
105         }
106     }
107     
108     void dfs(int p, int cur) {
109         F[cur].ok = 0;
110         deal(p, F[cur].b, F[cur].a);
111         deal(p, F[cur].c, F[cur].b);
112         deal(p, F[cur].a, F[cur].c);
113     }
114     
115     bool same(int s, int t) {
116         pt &a = P[F[s].a], &b = P[F[s].b], &c = P[F[s].c];
117         return fabs(volume(a, b, c, P[F[t].a])) < eps && fabs(volume(a, b, c, P[F[t].b])) < eps && fabs(volume(a, b, c, P[F[t].c])) < eps;
118     }
119     
120     //构建三维凸包
121     void construct() {
122         cnt = 0;
123         if (n < 4)
124             return;
125         
126         /*********此段是为了保证前四个点不公面，若已保证，可去掉********/
127         bool sb = 1;
128         //使前两点不公点
129         for (int i = 1; i < n; i++) {
130             if (vlen(P[0] - P[i]) > eps) {
131                 swap(P[1], P[i]);
132                 sb = 0;
133                 break;
134             }
135         }
136         if (sb)return;
137         
138         sb = 1;
139         //使前三点不公线
140         for (int i = 2; i < n; i++) {
141             if (vlen(Cross3((P[0] - P[1]) , (P[1] - P[i]))) > eps) {
142                 swap(P[2], P[i]);
143                 sb = 0;
144                 break;
145             }
146         }
147         if (sb)return;
148         
149         sb = 1;
150         //使前四点不共面
151         for (int i = 3; i < n; i++) {
152             if (fabs(Cross3((P[0] - P[1]) , (P[1] - P[2])) ^ (P[0] - P[i])) > eps) {
153                 swap(P[3], P[i]);
154                 sb = 0;
155                 break;
156             }
157         }
158         if (sb)return;
159         /*********此段是为了保证前四个点不公面********/
160         
161         
162         fac add;
163         for (int i = 0; i < 4; i++) {
164             add.a = (i+1)%4, add.b = (i+2)%4, add.c = (i+3)%4, add.ok = 1;
165             if (ptof(P[i], add) > 0)
166                 swap(add.b, add.c);
167             to[add.a][add.b] = to[add.b][add.c] = to[add.c][add.a] = cnt;
168             F[cnt++] = add;
169         }
170         
171         for (int i = 4; i < n; i++) {
172             for (int j = 0; j < cnt; j++) {
173                 if (F[j].ok && ptof(P[i], F[j]) > eps) {
174                     dfs(i, j);
175                     break;
176                 }
177             }
178         }
179         int tmp = cnt;
180         cnt = 0;
181         for (int i = 0; i < tmp; i++) {
182             if (F[i].ok) {
183                 F[cnt++] = F[i];
184             }
185         }
186     }
187     
188     //表面积
189     double area() {
190         double ret = 0.0;
191         for (int i = 0; i < cnt; i++) {
192             ret += area(P[F[i].a], P[F[i].b], P[F[i].c]);
193         }
194         return ret / 2.0;
195     }
196     
197     //体积
198     double volume() {
199         pt O(0, 0, 0);
200         double ret = 0.0;
201         for (int i = 0; i < cnt; i++) {
202             ret += volume(O, P[F[i].a], P[F[i].b], P[F[i].c]);
203         }
204         return fabs(ret / 6.0);
205     }
206     
207     //表面三角形数
208     int facetCnt_tri() {
209         return cnt;
210     }
211     
212     //表面多边形数
213     int facetCnt() {
214         int ans = 0;
215         for (int i = 0; i < cnt; i++) {
216             bool nb = 1;
217             for (int j = 0; j < i; j++) {
218                 if (same(i, j)) {
219                     nb = 0;
220                     break;
221                 }
222             }
223             ans += nb;
224         }
225         return ans;
226     }
227     pt centroid(){
228         pt ans(0,0,0),o(0,0,0);
229         double all=0;
230         for(int i=0;i<cnt;i++)
231         {
232             double vol=volume(o,P[F[i].a],P[F[i].b],P[F[i].c]);
233             ans=ans+(o+P[F[i].a]+P[F[i].b]+P[F[i].c])/4.0*vol;
234             all+=vol;
235         }
236         ans=ans/all;
237         return ans;
238     }
239     double res(){
240         pt a=centroid();
241         double _min=1e10;
242         for(int i=0;i<cnt;++i){
243             double now=Dis(P[F[i].a],P[F[i].b],P[F[i].c],a);
244             _min=min(_min,now);
245         }
246         return _min;
247     }
248     double ptoface(pt p,int i)
249     {
250         return fabs(volume(P[F[i].a],P[F[i].b],P[F[i].c],p)/vlen((P[F[i].b]-P[F[i].a])*(P[F[i].c]-P[F[i].a])));
251     }
252 }lou;
253 bool mult(pp sp,pp ep,pp op){
254     return (sp.x-op.x)*(ep.y-op.y)>=(ep.x-op.x)*(sp.y-op.y);
255 }
256 double Cross(pp a,pp b,pp c){
257     return (c.x-a.x)*(b.y-a.y) - (b.x-a.x)*(c.y-a.y);
258 }
259 
260 bool cmp(pp a,pp b){
261     if(a.y==b.y)return a.x<b.x;
262     return a.y<b.y;
263 }
264 int n,res[60],top;
265 pp ps[60];
266 void Graham(){
267     int len;
268     n=lou.n;
269     top=1;
270     sort(ps,ps+n,cmp);
271     if(n==0)return;res[0]=0;
272     if(n==1)return;res[1]=1;
273     if(n==2)return;res[2]=2;
274     for(int i=2;i<n;i++){
275         while(top&&mult(ps[i],ps[res[top]],ps[res[top-1]]))
276             top--;
277         res[++top]=i;
278     }
279     len=top;
280     res[++top]=n-2;
281     for(int i=n-3;i>=0;i--){
282         while(top!=len&&mult(ps[i],ps[res[top]],ps[res[top-1]]))top--;
283         res[++top]=i;
284     }
285 }
286 inline pt get_point(pt st,pt ed,pt tp)
287 {
288     double t1=(tp-st)^(ed-st);
289     double t2=(ed-st)^(ed-st);
290     double t=t1/t2;
291     pt ans=st + ((ed-st)*t);
292     return ans;
293 }
294 inline double dist(pt st,pt ed)
295 {
296     return sqrt((ed-st)^(ed-st));
297 }
298 pp rotate(pt st,pt ed,pt tp,double A)
299 {
300     pt root=get_point(st,ed,tp);
301     pt e=(ed-st)/dist(ed,st);
302     pt r=tp-root;
303     pt vec=e*r;
304     pt ans=r*cos(A)+vec*sin(A)+root;
305     return pp(ans.x,ans.y);
306 }
307 int main(){
308     while (scanf("%d",&lou.n)&&lou.n) {
309         for (int i=0; i<lou.n; ++i)
310             scanf("%lf%lf%lf",&lou.P[i].x,&lou.P[i].y,&lou.P[i].z);
311         lou.construct();
312         double ansh=0,ansa=inf;
313         if(lou.n<=2)
314         {
315             printf("0.000 0.000\n");
316         }
317         else if(lou.n==3)
318         {
319             ansh=0;
320             ansa=(lou.P[1]-lou.P[0])^(lou.P[2]-lou.P[0]);
321             ansa/=2.0;
322             printf("%.3lf %.3lf\n",ansh,ansa);
323             
324         }
325         else {
326             for (int i=0; i<lou.cnt; ++i) {
327                 pt p1=(lou.P[lou.F[i].b]-lou.P[lou.F[i].a])*(lou.P[lou.F[i].c]-lou.P[lou.F[i].a]);
328                 pt e=pt(0,0,1);
329                 pt vec=p1*e;
330                 double A=p1^e/p1.len();
331                 A=acos(A);
332                 if(fabs(A-pi)>EPS&&fabs(A)>EPS){
333                     pt s=pt(0,0,0);
334                     for (int k=0; k<lou.n; ++k) ps[k]=rotate(s,vec,lou.P[k],A);
335                 }
336                 else
337                 {
338                     for(int k=0; k<lou.n; ++k) ps[k].x=lou.P[k].x,ps[k].y=lou.P[k].y;
339                 }
340                 double h=0;
341                 for (int j=0; j<lou.n; ++j)
342                     h=max(lou.ptoface(lou.P[j],i),h);
343                 if (h<ansh)
344                     continue;
345                 Graham();
346                 double a=0;
347                 for (int k=1; k<top-1; ++k){
348                     a+=fabs(Cross(ps[res[0]],ps[res[k]],ps[res[k+1]]));
349                 }
350                 a/=2.0;
351                 if (fabs(h-ansh)<EPS) {
352                     ansa=min(ansa,a);
353                 }
354                 else
355                     ansa=a;
356                 ansh=h;
357             }
358             printf("%.3lf %.3lf\n",ansh,ansa);
359         }
360     }
361 }
View Code

 

Problem I

签到题，就是计算所有数的平方和

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstdlib>
 5 #include<algorithm>
 6 #include<cstring>
 7 #include<string>
 8 #include<vector>
 9 #include<map>
10 #include<set>
11 #include<queue>
12 using namespace std;
13 typedef long long ll;
14 ll n;
15 int main()
16 {
17     while (~scanf("%lld",&n))
18     {
19         if (n==0) break;
20         ll sum=0;
21         while (n--)
22         {
23             ll x;
24             scanf("%lld",&x);
25             sum+=x*x;
26         }
27         printf("%lld\n",sum);
28     }
29     return 0;
30 }
View Code

 

Problem J

由于只有裤子跟衣服或鞋子搭配不合法

因此可以求出每条裤子跟多少衣服和鞋子搭配是合法的

然后用乘法原理进行统计即可

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstdlib>
 5 #include<algorithm>
 6 #include<cstring>
 7 #include<string>
 8 #include<vector>
 9 #include<map>
10 #include<set>
11 #include<queue>
12 using namespace std;
13 typedef long long ll;
14 int n,m,k,sum1[1010],sum2[1010];
15 char s1[11],s2[11];
16 int main()
17 {
18     while (~scanf("%d%d%d",&n,&m,&k))
19     {
20         if (n==0&&m==0&&k==0) break;
21         int q;
22         int i;
23         for (i=1;i<=m;i++)
24         {
25             sum1[i]=n;
26             sum2[i]=k;
27         }
28         scanf("%d",&q);
29         while (q--)
30         {
31             int x,y;
32             scanf("%s%d%s%d",s1,&x,s2,&y);
33             if (s1[0]=='c') sum1[y]--;
34             else sum2[x]--;
35         }
36         ll ans=0;
37         for (i=1;i<=m;i++)
38             ans+=1ll*sum1[i]*sum2[i];
39         printf("%lld\n",ans);
40     }
41     return 0;
42 }
View Code

 

Problem K

根据题意进行模拟即可

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cmath>
  4 #include<cstdlib>
  5 #include<algorithm>
  6 #include<cstring>
  7 #include<string>
  8 #include<vector>
  9 #include<map>
 10 #include<set>
 11 #include<queue>
 12 using namespace std;
 13 #define y1 dfggf
 14 #define y2 kkljk
 15 const int d[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
 16 int c1,s1,t1,c2,s2,t2,n,t;
 17 int main()
 18 {
 19     while (~scanf("%d",&n))
 20     {
 21         if (n==0) break;
 22         char c[2];
 23         scanf("%s%d%d",c,&s1,&t1);
 24         if (c[0]=='E') c1=0;
 25         else if (c[0]=='S') c1=1;
 26         else if (c[0]=='W') c1=2;
 27         else c1=3;
 28         scanf("%s%d%d",c,&s2,&t2);
 29         if (c[0]=='E') c2=0;
 30         else if (c[0]=='S') c2=1;
 31         else if (c[0]=='W') c2=2;
 32         else c2=3;
 33         scanf("%d",&t);
 34         int i;
 35         int x1=1,y1=1,x2=n,y2=n;
 36         for (i=1;i<=t;i++)
 37         {
 38             int nx1=x1+d[c1][0]*s1;
 39             int ny1=y1+d[c1][1]*s1;
 40             if (nx1<1)
 41             {
 42                 nx1=1+(1-nx1);
 43                 c1=1;
 44             }
 45             if (nx1>n)
 46             {
 47                 nx1=n-(nx1-n);
 48                 c1=3;
 49             }
 50             if (ny1<1)
 51             {
 52                 ny1=1+(1-ny1);
 53                 c1=0;
 54             }
 55             if (ny1>n)
 56             {
 57                 ny1=n-(ny1-n);
 58                 c1=2;
 59             }
 60             int nx2=x2+d[c2][0]*s2;
 61             int ny2=y2+d[c2][1]*s2;
 62             if (nx2<1)
 63             {
 64                 nx2=1+(1-nx2);
 65                 c2=1;
 66             }
 67             if (nx2>n)
 68             {
 69                 nx2=n-(nx2-n);
 70                 c2=3;
 71             }
 72             if (ny2<1)
 73             {
 74                 ny2=1+(1-ny2);
 75                 c2=0;
 76             }
 77             if (ny2>n)
 78             {
 79                 ny2=n-(ny2-n);
 80                 c2=2;
 81             }
 82             if (nx1==nx2&&ny1==ny2)
 83                 swap(c1,c2);
 84             else
 85             {
 86                 if (i%t1==0)
 87                 {
 88                     c1--;
 89                     if (c1==-1) c1=3;
 90                 }
 91                 if (i%t2==0)
 92                 {
 93                     //cout<<c2<<" ";
 94                     c2--;
 95                     //cout<<i<<" "<<c2<<endl;
 96                     if (c2==-1) c2=3;
 97                 }
 98             }
 99             x1=nx1;
100             y1=ny1;
101             x2=nx2;
102             y2=ny2;    
103             //cout<<i<<endl;
104             //cout<<x1<<" "<<y1<<" "<<c1<<endl;
105             //cout<<x2<<" "<<y2<<" "<<c2<<endl;
106         }
107         printf("%d %d\n",x1,y1);
108         printf("%d %d\n",x2,y2);
109     }
110     return 0;
111 }
View Code

 

posted @ 2017-10-04 18:46  cxhscst2  阅读(371)  评论(0)  编辑  收藏  举报