2017.2其他简要题解

codeforces 739A

答案一定是所有区间中最小区间长度，构造的话就把所有区间都当成那个长度即可

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 int b[100010];
 5 struct node{int x,y;} a[100010];
 6 int n,m;
 7 
 8 bool cmp(node a,node b)
 9 {
10     return a.x<b.x;
11 }
12 
13 int main()
14 {
15     scanf("%d%d",&n,&m);
16     int ans=n;
17     for (int i=1; i<=m; i++)
18     {
19         scanf("%d%d",&a[i].x,&a[i].y);
20         ans=min(ans,a[i].y-a[i].x+1);
21     }
22     sort(a+1,a+1+m,cmp);
23     for (int i=1; i<=m; i++)
24         a[i].y=a[i].x+ans-1;
25     printf("%d\n",ans);
26     for (int i=1; i<=m; i++)
27     {
28         if (a[i].x==a[i-1].x) continue;
29         if (a[i].x>a[i-1].y)
30         {
31             for (int j=0; j<ans; j++)
32                 b[a[i].x+j]=j;
33         }
34         else {
35             int k=a[i-1].y+1;
36             for (int j=a[i-1].x; j<=a[i].x; j++)
37                 b[k++]=b[j];
38         }
39     }
40     for (int i=1; i<=n; i++) printf("%d ",b[i]);
41 }

View Code

codeforces 739B

做法很多，可以树上差分统计然后二分点到根这段，也可以直接用可持久化线段树维护

 1 #include<bits/stdc++.h>
 2 #define mp make_pair
 3 using namespace std;
 4 typedef long long ll;
 5 struct way{int po,next,len;} e[200010];
 6 int ans[200010],p[200010],a[200010];
 7 ll d[200010];
 8 vector< pair<ll,int> > st;
 9 int n,len;
10 
11 void add(int x,int y,int z)
12 {
13     e[++len].po=y;
14     e[len].next=p[x];
15     e[len].len=z;
16     p[x]=len;
17 }
18 
19 void dfs(int x)
20 {
21     st.push_back(mp(d[x],x));
22     int j=lower_bound(st.begin(),st.end(),mp(d[x]-a[x],0))-st.begin()-1;
23     if (j>=0) ans[st[j].second]--;
24     for (int i=p[x]; i; i=e[i].next)
25     {
26         int y=e[i].po;
27         d[y]=d[x]+e[i].len;
28         dfs(y);
29         ans[x]+=ans[y]+1;
30     }
31     st.pop_back();
32 }
33 
34 int main()
35 {
36     scanf("%d",&n);
37     for (int i=1; i<=n; i++) scanf("%d",&a[i]);
38     for (int i=2; i<=n; i++)
39     {
40         int x,y;
41         scanf("%d%d",&x,&y);
42         add(x,i,y);
43     }
44     dfs(1);
45     for (int i=1; i<=n; i++) printf("%d ",ans[i]);
46 }

View Code

hdu5071

vector大模拟

  1 #include<bits/stdc++.h>
  2 
  3 using namespace std;
  4 #define mp make_pair
  5 
  6 typedef long long ll;
  7 typedef pair<int,ll> pi;
  8 vector<pi> p;
  9 int x,now;
 10 
 11 int get(int x)
 12 {
 13     for (int i=0; i<p.size(); i++)
 14         if (p[i].first==x) return i;
 15     return -1;
 16 }
 17 
 18 void erase(int x)
 19 {
 20     int i=0;
 21     for( vector<pi>::iterator it=p.begin();it!=p.end(); it++,i++){
 22         if (i==x)
 23         {
 24             p.erase(it);
 25             return;
 26         }
 27     }
 28 }
 29 
 30 void add()
 31 {
 32     scanf("%d",&x);
 33     if (get(x)!=-1) puts("same priority.");
 34     else {
 35         p.push_back(mp(x,0));
 36         puts("success.");
 37     }
 38 }
 39 
 40 void close()
 41 {
 42     scanf("%d",&x);
 43     int k=get(x);
 44     if (k==-1) puts("invalid priority.");
 45     else{
 46         if(x==now) now=0;
 47         printf("close %d with %I64d.\n",p[k].first,p[k].second);
 48         erase(k);
 49     }
 50 }
 51 void chat()
 52 {
 53     scanf("%d",&x);
 54     if (!p.size()) puts("empty.");
 55     else {
 56         if (now) p[get(now)].second+=x;
 57         else p[0].second+=x;
 58         puts("success.");
 59     }
 60 }
 61 
 62 void rot(int y)
 63 {
 64     if (y<1||y>p.size()) puts("out of range.");
 65     else{
 66         pi pp=p[y-1];
 67         erase(y-1);
 68         p.insert(p.begin(),pp);
 69         puts("success.");
 70     }
 71 }
 72 
 73 void prior()
 74 {
 75     int mx=0,k=-1;
 76     if (!p.size()) puts("empty.");
 77     else {
 78         for(int i=0;i<p.size();++i)
 79             if(p[i].first>mx) {mx=p[i].first;k=i;}
 80         rot(k+1);
 81     }
 82 }
 83 void choose()
 84 {
 85     scanf("%d",&x);
 86     int ch=get(x);
 87     if (ch==-1) puts("invalid priority.");
 88     else rot(ch+1);
 89 }
 90 
 91 void top()
 92 {
 93     scanf("%d",&x);
 94     if(get(x)==-1) puts("invalid priority.");
 95     else now=x,puts("success.");
 96 }
 97 void untop()
 98 {
 99     if (now) {now=0;puts("success.");}
100     else puts("no such person.");
101 }
102 
103 void bye()
104 {
105     if (now)
106     {
107         x=get(now);
108         if(p[x].second!=0) printf("Bye %d: %I64d\n",p[x].first,p[x].second);
109         erase(x);
110     }
111     for (int i=0; i<p.size(); i++)
112         if (p[i].second) printf("Bye %d: %I64d\n",p[i].first,p[i].second);
113 }
114 int main()
115 {
116     string op;
117     int cas,y,n; scanf("%d",&cas);
118     while(cas--)
119     {
120         now=0;
121         scanf("%d",&n);
122         p.clear();
123         for(int k=1; k<=n; k++)
124         {
125             cin>>op;
126             printf("Operation #%d: ",k);
127             if(op=="Add") add();
128             else if(op=="Close") close();
129             else if(op=="Chat") chat();
130             else if(op=="Rotate") {scanf("%d",&y);rot(y);}
131             else if(op=="Prior") prior();
132             else if(op=="Choose") choose();
133             else if(op=="Top") top();
134             else if(op=="Untop") untop();
135         }
136         bye();
137     }
138     return 0;
139 }

View Code

hdu5073

方差贪心，最后答案一定是连续的n-k个组成的惯量，注意精度保护

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 typedef long long ll;
 5 int n,k,a[100010];
 6 int main()
 7 {
 8     int cas;
 9     scanf("%d",&cas);
10     while (cas--)
11     {
12         scanf("%d%d",&n,&k);
13         for (int i=1; i<=n; i++)
14             scanf("%d",&a[i]);
15         sort(a+1,a+1+n);
16         if (k>=n-1)
17         {
18             puts("0");
19             continue;
20         }
21         ll s1=0,s2=0;
22         for (int i=1; i<=n-k; i++)
23         {
24             s2+=1ll*a[i]*a[i];
25             s1+=a[i];
26         }
27         ll ans=1ll*(n-k)*s2-s1*s1;
28         for (int i=n-k+1; i<=n; i++)
29         {
30             s2+=1ll*a[i]*a[i]-1ll*a[i-n+k]*a[i-n+k];
31             s1+=a[i]-a[i-n+k];
32             ans=min(ans,1ll*(n-k)*s2-s1*s1);
33         }
34         printf("%.10lf\n",1.0*ans/(n-k));
35     }
36 }

View Code

hdu5074

简单dp

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 const int inf=-1000000007;
 5 int f[2][60],a[60][60],n,m;
 6 int main()
 7 {
 8     int cas;
 9     scanf("%d",&cas);
10     while (cas--)
11     {
12         scanf("%d%d",&n,&m);
13         for (int i=1; i<=m; i++)
14             for (int j=1; j<=m; j++)
15                 scanf("%d",&a[i][j]);
16         for (int i=1; i<=m; i++) f[0][i]=inf;
17         int x;
18         scanf("%d",&x);
19         if (x==-1) memset(f[0],0,sizeof(f[0])); else f[0][x]=0;
20         int p=0;
21         for (int i=2; i<=n; i++)
22         {
23             p^=1;
24             for (int j=1; j<=m; j++) f[p][j]=inf;
25             scanf("%d",&x);
26             if (x>-1)
27             {
28                 for (int j=1; j<=m; j++)
29                     f[p][x]=max(f[p][x],f[p^1][j]+a[j][x]);
30             }
31             else {
32                 for (int j=1; j<=m; j++)
33                     for (int k=1; k<=m; k++)
34                         f[p][k]=max(f[p][k],f[p^1][j]+a[j][k]);
35             }
36         }
37         int ans=0;
38         for (int i=1; i<=m; i++) ans=max(ans,f[p][i]);
39         printf("%d\n",ans);
40     }
41 }

View Code

hdu5078

水题简单模拟

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 double sqr(double x)
 6 {
 7     return x*x;
 8 }
 9 
10 double dis(int x,int y,int x0,int y0)
11 {
12     return sqrt(sqr(x-x0)+sqr(y-y0));
13 }
14 
15 int main()
16 {
17     int cas,n;
18     scanf("%d",&cas);
19     while (cas--)
20     {
21         scanf("%d",&n);
22         double ans=0;
23         int x0,y0,x,y,t,t0;
24         scanf("%d%d%d",&t0,&x0,&y0);
25         for (int i=2; i<=n; i++)
26         {
27             scanf("%d%d%d",&t,&x,&y);
28             ans=max(ans,dis(x,y,x0,y0)/(t-t0));
29             x0=x,y0=y,t0=t;
30         }
31         printf("%.10lf\n",ans);
32     }
33 }

View Code

hdu5077

暴力dfs+剪枝，然后打表

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 const int ans[300]={1,5,6,3,6,3,7,4,7,8,4,5,4,5,4,1,6,3,7,4,7,4,9,7,8,8,7,5,7,5,7,4,7,8,4,5,8,8,7,5,8,9,5,6,8,8,5,5,4,5,4,1,7,5,7,4,8,8,5,5,5,7,6,4,7,8,8,8,4,5,7,5,8,9,8,8,5,6,5,5,4,5,7,5,4,1,7,4,8,8,5,7,5,5,6,4,8,9,8,8,8,8,5,7,11,9,8,9,8,9,8,8,5,6,5,5,5,5,6,4,8,9,8,8,8,8,8,7,8,9,9,9,9,9,10,9,5,7,6,6,6,6,7,6,9,9,10,9,10,9,10,10,7,6,7,7,7,7,9,7,5,7,6,6,7,6,7,7,5,6,2,3,6,6,4,3,6,6,7,6,7,7,9,7,6,6,4,3,7,7,7,6,5,7,7,6,6,6,7,7,5,6,6,6,2,3,4,3,6,6,7,7,7,6,9,7,6,6,7,7,4,3,7,6,5,6,6,6,6,6,7,7,8,9,5,6,5,6,2,5,2,3,4,3,4,3,7,6,5,6,2,5,2,5,4,1};
 6 
 7 int main()
 8 {
 9     int cas;
10     char s[10];
11     scanf("%d",&cas);
12     while (cas--)
13     {
14         scanf("%s",s);
15         int n=0;
16         for (int i=0; i<8; i++)
17         {
18             n<<=1;
19             n|=s[i]-'0';
20         }
21         printf("%d\n",ans[n]);
22     }
23 }

View Code

hdu5860

设编号0~n-1,a[i]为第i轮死，则a[i]=(i%k)?a[i-i/k-1]+1:1，然后整理一下就好

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 int a[3000010],ans[3000010],g[3000010],nex[3000010],p[3000010],len,n,m,q;
 6 void add(int x,int y)
 7 {
 8     g[++len]=y;
 9     nex[len]=p[x];
10     p[x]=len;
11 }
12 
13 int main()
14 {
15     int cas;
16     scanf("%d",&cas);
17     while (cas--)
18     {
19         scanf("%d%d%d",&n,&m,&q);
20         len=0; int mx=1;
21         for (int i=0; i<n; i++)
22         {
23             if (i%m) a[i]=a[i-i/m-1]+1;
24             else a[i]=1;
25             mx=max(a[i],mx);
26         }
27         for (int i=n-1; i>=0; i--) add(a[i],i+1);
28         int t=0;
29         for (int i=1; i<=mx; i++)
30             for (int j=p[i]; j; j=nex[j]) ans[++t]=g[j];
31         while (q--)
32         {
33             int x; scanf("%d",&x);
34             printf("%d\n",ans[x]);
35         }
36         for (int i=1; i<=mx; i++) p[i]=0;
37     }
38 }

View Code

hdu5861

维护每条线段最早和最迟用到为第几天，然后差分计算答案

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 int ans[200010],mx[800010],mi[800010],w[200010],n,m;
 5 void push(int i)
 6 {
 7     if (mi[i]<=m)
 8     {
 9         mi[i*2]=min(mi[i*2],mi[i]);
10         mi[i*2+1]=min(mi[i*2+1],mi[i]);
11         mi[i]=m+1;
12     }
13     if (mx[i])
14     {
15         mx[i*2]=max(mx[i*2],mx[i]);
16         mx[i*2+1]=max(mx[i*2+1],mx[i]);
17         mx[i]=0;
18     }
19 }
20 
21 void build(int i,int l,int r)
22 {
23     mx[i]=0; mi[i]=m+1;
24     if (l!=r)
25     {
26         int m=(l+r)>>1;
27         build(i*2,l,m);
28         build(i*2+1,m+1,r);
29     }
30 }
31 
32 void add(int i,int l,int r,int x,int y,int z)
33 {
34     if (x<=l&&y>=r)
35     {
36         mi[i]=min(mi[i],z);
37         mx[i]=max(mx[i],z);
38     }
39     else {
40         int m=(l+r)>>1;
41         push(i);
42         if (x<=m) add(i*2,l,m,x,y,z);
43         if (y>m) add(i*2+1,m+1,r,x,y,z);
44     }
45 }
46 
47 void ask(int i,int l,int r)
48 {
49     if (l==r)
50     {
51         if (mx[i]>=mi[i])
52         {
53             ans[mi[i]]+=w[l];
54             ans[mx[i]+1]-=w[l];
55         }
56     }
57     else {
58         int m=(l+r)>>1;
59         push(i);
60         ask(i*2,l,m);
61         ask(i*2+1,m+1,r);
62     }
63 }
64 
65 int main()
66 {
67     while (scanf("%d%d",&n,&m)!=EOF)
68     {
69         for (int i=1; i<n; i++) scanf("%d",&w[i]);
70         memset(ans,0,sizeof(ans));
71         if (n-1) build(1,1,n-1);
72         for (int i=1; i<=m; i++)
73         {
74             int x,y;
75             scanf("%d%d",&x,&y);
76             if (x>y) swap(x,y);
77             if (n-1) add(1,1,n-1,x,y-1,i);
78         }
79         if (n-1) ask(1,1,n-1);
80         for (int i=1; i<=m; i++) ans[i]+=ans[i-1];
81         for (int i=1; i<=m; i++) printf("%d\n",ans[i]);
82     }
83 }

View Code

hdu5862

只有可能竖直线段和横向线段有交点，那么就是经典的扫描线了：x排序，维护y，竖直线段查询，横向线段“头进尾出”

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 typedef long long ll;
 5 struct node{int x,l,r,id;} a[200010];
 6 int tr[200010*4],b[200010],n;
 7 
 8 bool cmp(node a,node b)
 9 {
10     if (a.x==b.x) return a.id<b.id;
11     return a.x<b.x;
12 }
13 
14 void build(int i,int l,int r)
15 {
16     tr[i]=0;
17     if (l!=r)
18     {
19         int m=(l+r)>>1;
20         build(i*2,l,m);
21         build(i*2+1,m+1,r);
22     }
23 }
24 
25 void add(int i,int l,int r,int x,int y)
26 {
27     if (l==r) tr[i]+=y;
28     else {
29         int m=(l+r)>>1;
30         if (x<=m) add(i*2,l,m,x,y);
31         else add(i*2+1,m+1,r,x,y);
32         tr[i]=tr[i*2]+tr[i*2+1];
33     }
34 }
35 
36 int ask(int i,int l,int r,int x,int y)
37 {
38     if (x<=l&&y>=r) return tr[i];
39     else {
40         int m=(l+r)>>1,s=0;
41         if (x<=m) s+=ask(i*2,l,m,x,y);
42         if (y>m) s+=ask(i*2+1,m+1,r,x,y);
43         return s;
44     }
45 }
46 
47 int main()
48 {
49     int cas;
50     scanf("%d",&cas);
51     while (cas--)
52     {
53         scanf("%d",&n);
54         int t=0,m=0;
55         for (int i=1; i<=n; i++)
56         {
57             int x1,y1,x2,y2;
58             scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
59             b[++m]=y1; b[++m]=y2;
60             if (x1==x2)
61             {
62                 a[++t].x=x1;
63                 if (y1>y2) swap(y1,y2);
64                 a[t].l=y1; a[t].r=y2;
65                 a[t].id=2;
66             }
67             else {
68                 if (x1>x2) swap(x1,x2);
69                 a[++t].x=x1;
70                 a[t].l=a[t].r=y1; a[t].id=1;
71                 a[++t].x=x2+1;
72                 a[t].l=a[t].r=y1; a[t].id=-1;
73             }
74         }
75         sort(a+1,a+1+t,cmp);
76         sort(b+1,b+1+m);
77         m=unique(b+1,b+m+1)-b-1;
78         for (int i=1; i<=t; i++)
79         {
80             a[i].l=lower_bound(b+1,b+1+m,a[i].l)-b;
81             if (a[i].id==2) a[i].r=lower_bound(b+1,b+1+m,a[i].r)-b;
82         }
83         build(1,1,m);
84         ll ans=0;
85         for (int i=1; i<=t; i++)
86             if (a[i].id==2) ans+=ask(1,1,m,a[i].l,a[i].r);
87             else add(1,1,m,a[i].l,a[i].id);
88 
89         printf("%lld\n",ans);
90     }
91 }

View Code

hdu5961

把P图建出来找出环则为N，再把Q图反向找出环也为N

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 int d[2050],q[2050],n;
 5 vector<int> g[2050];
 6 pair<int,int> w[2200000];
 7 char s[2050];
 8 
 9 bool cycle()
10 {
11     int f=1,r=0;
12     memset(d,0,sizeof(d));
13     for (int i=1; i<=n; i++)
14         for (int j=0; j<g[i].size(); j++) d[g[i][j]]++;
15     for (int i=1; i<=n; i++)
16         if (!d[i]) q[++r]=i;
17     while (f<=r)
18     {
19         int x=q[f++];
20         for (int i=0; i<g[x].size(); i++)
21         {
22             int y=g[x][i];
23             if (--d[y]==0) q[++r]=y;
24         }
25     }
26     return r<n;
27 }
28 
29 int main()
30 {
31     int cas;
32     scanf("%d",&cas);
33     while (cas--)
34     {
35         scanf("%d",&n);
36         for (int i=1; i<=n; i++) g[i].clear();
37         int t=0;
38         for (int i=1; i<=n; i++)
39         {
40             scanf("%s",s+1);
41             for (int j=1; j<=n; j++)
42                 if (s[j]=='P') g[i].push_back(j);
43                 else if (s[j]=='Q') w[++t]=make_pair(i,j);
44         }
45         if (cycle()) {puts("N"); continue;}
46         else {
47             for (int i=1; i<=t; i++)
48                 g[w[i].second].push_back(w[i].first);
49             if (cycle()) puts("N");
50             else puts("T");
51         }
52     }
53 }

View Code

hdu5802

尽可能的往下，处理一下停顿就好

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 int p,q;
 5 
 6 int dfs(int p,int ans,int s)
 7 {
 8     if (p==q) return ans;
 9     int x=0;
10     while (p-(1<<x)+1>q) x++;
11     if (p-(1<<x)+1==q) return ans+x;
12     int ans0=ans+x+max(0,q-max(0,p+1-(1<<x))-s);
13     return min(ans0,dfs(p-(1<<(x-1))+1,ans+x,s+1));
14 }
15 
16 int main()
17 {
18     int cas;
19     scanf("%d",&cas);
20     while (cas--)
21     {
22         scanf("%d%d",&p,&q);
23         if (q>=p) printf("%d\n",q-p);
24         else printf("%d\n",dfs(p,0,0));
25     }
26 }

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posted on 2017-02-03 22:34 acphile 阅读(189) 评论(0) 编辑收藏举报