今天讲了计算几何，发几道水水的tyvj上的题解...
计算几何好难啊！@Mrs.General....怎么办....
这几道题都是在省选之前做的，所以前面的Point运算啊，dcmp啊，什么什么的，基本上没用，每次都把上次的main()函数删了接着继续写....
原谅我曾经丑出翔的代码...
虽然现在也很丑...
TYVJ 1150 绳子围点
题解：
凸包 + 凸包面积 + 皮克定理
1 #include <cstdio>
2 #include <algorithm>
3 using namespace std;
4 typedef long long ll;
5
6 const int MAXN = 200000 + 10;
7
8 char cj;
9 int n, m;
10
11 struct Point{//{{{
12 ll x, y;
13
14 friend bool operator < (const Point& A, const Point& B){
15 return A.x < B.x || (A.x == B.x && A.y < B.y);
16 }
17 } p[MAXN], ch[MAXN * 2];//}}}
18
19 inline ll xmult(Point a, Point b, Point c){//{{{
20 return (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y);
21 }
22
23 inline void AndrweScan(){
24 sort(p, p + n);
25 m = 1, ch[0] = p[0], ch[1] = p[1];
26 for (int i = 2; i < n; i ++){
27 while (m >= 1 && xmult(ch[m], ch[m - 1], p[i]) > 0)
28 m --;
29 ch[++ m] = p[i];
30 }
31
32 for (int i = n - 2; i >= 0; i --){
33 while (m >= 1 && xmult(ch[m], ch[m - 1], p[i]) > 0)
34 m --;
35 ch[++ m] = p[i];
36 }
37 }//}}}
38
39 inline ll nextLL(){
40 bool flag = false;
41 do {
42 cj = getchar();
43 if (cj == '-')
44 flag = true;
45 } while (!(48 <= cj && cj <= 57));
46
47 ll ret = 0;
48 do {
49 ret = ret * 10 + cj - 48;
50 cj = getchar();
51 } while (48 <= cj && cj <= 57);
52
53 return flag ? -ret : ret;
54 }
55
56 inline int nextInt(){
57 do
58 cj = getchar();
59 while (!(48 <= cj && cj <= 57));
60
61 int ret = 0;
62 do {
63 ret = ret * 10 + cj - 48;
64 cj = getchar();
65 } while (48 <= cj && cj <= 57);
66
67 return ret;
68 }
69
70 inline ll Plot(){
71 ll ret = 0;
72 for (int i = 2; i < m; i ++)
73 ret += xmult(ch[0], ch[i - 1], ch[i]);
74 return ret;
75 }
76
77 inline ll gcd(ll a, ll b){
78 return b ? gcd(b, a % b) : a;
79 }
80
81 inline ll Cal(int i){
82 int j = (i == m - 1) ? 0 : i + 1,
83 dx = ch[i].x - ch[j].x, dy = ch[i].y - ch[j].y;
84 return gcd(abs(dx), abs(dy));
85 }
86
87 inline void Pick(){
88 ll dS = Plot(), b = 0, a;
89 for (int i = 0; i < m; i++)
90 b += Cal(i);
91 a = (dS - b + 2) / 2;
92 printf("%lld\n", a + b);
93 }
94
95 int main(){
96 n = nextInt();
97 for (int i = 0; i < n; i++)
98 p[i].x = nextLL(), p[i].y = nextLL();
99 AndrweScan();
100 Pick();
101 }
TYVJ 1543 房间最短路
题解：
Floyd，dp[i][j]表示到第i面墙的第j个点的最短路，注意在更新最小值的时候判断两个点的连通
1 #include <cstdio>
2 #include <cmath>
3 #include <cstring>
4 #include <algorithm>
5 using namespace std;
6 #define INF 100000000
7 #define eps 1e-9
8
9 const int MAXN = 1000;
10
11 int n;
12 double dp[MAXN][MAXN];
13
14 struct Point{
15 double x, y;
16
17 friend Point operator + (const Point A, const Point B){
18 return (Point){A.x+B.x, A.y+B.y};
19 }
20
21 friend Point operator - (const Point A, const Point B){
22 return (Point){A.x-B.x, A.y-B.y};
23 }
24
25 friend Point operator * (const Point A, const Point B){
26 return (Point){A.x*B.x, A.y*B.y};
27 }
28
29 friend Point operator / (const Point A, const Point B){
30 return (Point){A.x/B.x, A.y/B.y};
31 }
32 }p[MAXN][MAXN];
33
34 inline Point make_Point(double X, double Y){
35 return (Point){X, Y};
36 }
37
38 inline double len(Point A){
39 return sqrt(A.x*A.x+A.y*A.y);
40 }
41
42 inline bool In_region(double eg, double l, double r){
43 if (l>r) swap(l, r);
44 return (l-eps<=eg && eg<=r+eps);
45 }
46
47 inline bool Connected(int I1, int I2, int I3, int I4){
48 if (I1>I3) swap(I1, I3);
49 if ((I1==I3 && I2==I4) || I3-I1==1) return true;
50 double k = (p[I1][I2].y-p[I3][I4].y)/(p[I1][I2].x-p[I3][I4].x), b = p[I1][I2].y-p[I1][I2].x*k, bn;
51 for (int i=I1+1; i<I3; i++){
52 bn = p[i][0].x*k+b;
53 if (!In_region(bn, p[i][0].y, p[i][1].y) && !In_region(bn, p[i][2].y, p[i][3].y)) return false;
54 }
55 return true;
56 }
57
58 int main(){
59 scanf("%d", &n);
60
61 p[0][0] = p[0][1] = p[0][2] = p[0][3] = make_Point(0, 5);
62 p[n+1][0] = p[n+1][1] = p[n+1][2] = p[n+1][3] = make_Point(10, 5);
63
64 double x, in, apl;
65 for (int i=1; i<=n; i++){
66 scanf("%lf", &x);
67 for (int j=0; j<4; j++)
68 scanf("%lf", &in), p[i][j] = make_Point(x, in);
69 }
70
71 //init
72 for (int i=0; i<4; i++)
73 dp[0][i] = 0.0;
74 for (int i=1; i<=n+1; i++)
75 for (int j=0; j<4; j++)
76 dp[i][j] = Connected(0, 0, i, j) ? len(p[i][j]-p[0][0]) : INF;
77
78 for (int i=0; i<=n+1; i++)
79 for (int j=0; j<4; j++)
80 for (int k=0; k<i; k++)
81 for (int l=0; l<4; l++)
82 if (Connected(k, l, i, j)) dp[i][j] = min(dp[i][j], dp[k][l]+len(p[i][j]-p[k][l]));
83 apl = INF;
84 for (int i=0; i<4; i++)
85 apl = min(apl, dp[n+1][i]);
86 printf("%.2lf\n",apl);
87 return 0;
88 }
TYVJ 1462 凸多边形
1 #include <cstdio>
2 #include <iomanip>
3 #include <iostream>
4 #include <algorithm>
5 #define eps 1e-12
6 using namespace std;
7
8 const int MAXN = 100001+10;
9
10 int n;
11
12 struct Point{
13 long double x, y;
14
15 friend Point operator - (const Point& A, const Point& B){
16 return (Point){A.x-B.x, A.y-B.y};
17 }
18
19 friend long double operator * (const Point& A, const Point& B){//cross
20 return A.x*B.y-A.y*B.x;
21 }
22
23 friend bool operator < (const Point& A, const Point& B){
24 return (A.x!=B.x) ? A.x<B.x : A.y<B.y;
25 }
26
27 inline void print(){
28 cout<<setiosflags(ios::fixed)<<setprecision(4)<<x<<" "<<setiosflags(ios::fixed)<<setprecision(4)<<y<<"\n";
29 }
30 }p[MAXN], ch[MAXN];
31
32 inline int dcmp(long double eg){
33 if (eg>eps) return 1; //>0
34 if (eg<-eps) return -1;//<0
35 else return 0;
36 }
37
38 inline void GrahamScan(){
39 int tot, t;
40 sort(p, p+n);
41
42 ch[0] = p[0], ch[1] = p[1], tot = 1;
43 for (int i=2; i<n; i++){
44 while (tot>0 && dcmp((ch[tot]-ch[tot-1])*(p[i]-ch[tot-1]))!=-1) tot--;
45 ch[++tot] = p[i];
46 }
47
48 t = tot, ch[++tot] = p[n-2];
49 for (int i=n-3; i>=0; i--){
50 while (tot>t && dcmp((ch[tot]-ch[tot-1])*(p[i]-ch[tot-1]))!=-1) tot--;
51 ch[++tot] = p[i];
52 }
53
54 for (int i=0; i<tot; i++)
55 ch[i].print();
56 }
57
58 int main(){
59 scanf("%d", &n);
60 for (int i=0; i<n; i++)
61 cin>>p[i].x>>p[i].y;
62
63 GrahamScan();
64 return 0;
65 }
TYVJ 1523 神秘大三角
题解：
这道题考的是读入...面积随便整一下就好了
1 #include <cstdio>
2 #include <cmath>
3 #include <algorithm>
4 #define eps 1e-9
5 using namespace std;
6
7 const int MAXL = 1000;
8
9 struct Point{
10 double x,y;
11
12 inline void in(){
13 char s[MAXL];
14 scanf("%s",s);
15 int i;
16 x = 0,y = 0;
17 for (i=1;;i++){
18 if (s[i]==',') break;
19 x *= 10,x += s[i]-48;
20 }
21 for (i+=1;;i++){
22 if (s[i]==')') break;
23 y *= 10,y += s[i]-48;
24 }
25
26 }
27
28 inline void out(){
29 printf("%.2lf %.2lf\n",x,y);
30 }
31
32 inline bool is_zero(){
33 return (y==0 && x==0);
34 }
35
36 inline double len(){
37 return sqrt(x*x-y*y);
38 }
39
40 friend Point operator + (const Point A,const Point B){
41 return (Point){A.x+B.x,A.y+B.y};
42 }
43
44 friend Point operator - (const Point A,const Point B){
45 return (Point){A.x-B.x,A.y-B.y};
46 }
47
48 friend Point operator * (const Point A,const double B){
49 return (Point){A.x*B,A.y*B};
50 }
51
52 friend Point operator / (const Point A,const double B){
53 return (Point){A.x/B,A.y/B};
54 }
55
56 friend bool operator == (const Point A,const Point B){
57 return (A.x==B.x && A.y==B.y);
58 }
59 }A,B,C,D;
60
61 inline double area(Point p1,Point p2,Point p3){
62 return abs(0.5*(p1.x*p2.y+p2.x*p3.y+p3.x*p1.y-p1.y*p2.x-p2.y*p3.x-p3.y*p1.x));
63 }
64
65 inline bool on(Point Aa,Point Ba,Point q){
66 double MX = max(Aa.x,Ba.x);
67 double mx = min(Aa.x,Ba.x);
68 double MY = max(Aa.y,Ba.y);
69 double my = min(Aa.y,Ba.y);
70 return ((mx<q.x && q.x<MX) && (my<q.y && q.y<MY)) ? true : false;
71 }
72
73 inline bool zero(double eg){
74 return (eg>-eps && eg<eps);
75 }
76
77 inline int appleeeeeeee(){
78 A.in();B.in();C.in();D.in();
79 if (A==D || B==D || C==D) return 4;//顶点上
80 double s1,s2,s3,s;
81 s = area(A,B,C);
82 s1 = area(A,B,D);
83 s2 = area(A,C,D);
84 s3 = area(B,C,D);
85
86 if (zero(s1) && on(A,B,D)) return 3;
87 if (zero(s2) && on(A,C,D)) return 3;
88 if (zero(s3) && on(B,C,D)) return 3;//边上
89
90 if ((s1+s2+s3-s)<eps && (s1+s2+s3-s)>-eps) return 1;
91 else return 2;
92 }
93
94 int main(){
95 printf("%d",appleeeeeeee());
96 return 0;
97 }
TYVJ 1544 角平分线
题解：
推个公式就好了
1 #include <cstdio>
2 #include <cmath>
3
4 struct Point{
5 double x,y;
6
7 inline void in(){
8 scanf("%lf%lf",&x,&y);
9 }
10
11 inline void out(){
12 printf("%.2lf %.2lf",x,y);
13 }
14
15 friend Point operator + (const Point A,const Point B){
16 return (Point){A.x+B.x,A.y+B.y};
17 }
18
19 friend Point operator - (const Point A,const Point B){
20 return (Point){A.x-B.x,A.y-B.y};
21 }
22
23 friend Point operator * (const Point A,const double B){
24 return (Point){A.x*B,A.y*B};
25 }
26
27 friend Point operator / (const Point A,const double B){
28 return (Point){A.x/B,A.y/B};
29 }
30 }a,b,c,d;
31
32 inline double len(Point eg){
33 return sqrt(eg.x*eg.x+eg.y*eg.y);
34 }
35
36 int main(){
37 a.in();b.in();c.in();
38 b = b-a,c = c-a;
39 double AB = len(b),AC = len(c);
40 double cj = (AB*AC)/(AB+AC);
41 d = b/AB + c/AC;
42 d = d*cj;
43 d = d + a;
44 d.out();
45 return 0;
46 }
