POJ2684 Polygonal Line Search 计算几何基础

　　属于比较水的计算几何题目，直接叉积判断即可。
 1 //STATUS:C++_AC_0MS_192KB
 2 #include<stdio.h>
 3 #include<stdlib.h>
 4 #include<string.h>
 5 #include<math.h>
 6 #include<iostream>
 7 #include<string>
 8 #include<algorithm>
 9 #include<vector>
10 #include<queue>
11 #include<stack>
12 using namespace std;
13 #define LL __int64
14 #define pii pair<int,int>
15 #define Max(a,b) ((a)>(b)?(a):(b))
16 #define Min(a,b) ((a)<(b)?(a):(b))
17 #define mem(a,b) memset(a,b,sizeof(a))
18 #define lson l,mid,rt<<1
19 #define rson mid+1,r,rt<<1|1
20 const int N=55,INF=0x3f3f3f3f,MOD=100000000;
21 const double DNF=100000000000;
22 
23 struct Node{
24     double x,y;
25 }nod[N][12];
26 
27 int n,cou[N];
28 
29 double dist(Node &a)
30 {
31     return a.x*a.x+a.y*a.y;
32 }
33 
34 void getr(Node& r,Node& a,Node& b)
35 {
36     r.x=a.x-b.x;
37     r.y=a.y-b.y;
38 }
39 
40 double cha(Node& r1,Node& r2)
41 {
42     return r1.x*r2.y-r2.x*r1.y;
43 }
44 
45 int judge(int k,int sta,int dir)
46 {
47     int i,j;
48     Node r0,rj,r01,rj1;
49     for(i=1,j=sta+dir;i<cou[0];i++,j+=dir){
50         getr(r0,nod[0][i],nod[0][i-1]);
51         getr(rj,nod[k][j],nod[k][j-dir]);
52         if(dist(r0)!=dist(rj))return 0;
53         if(i<cou[0]-1){
54             getr(r01,nod[0][i+1],nod[0][i]);
55             getr(rj1,nod[k][j+dir],nod[k][j]);
56             if(cha(r01,r0)!=cha(rj1,rj))return 0;
57         }
58     }
59     return 1;
60 }
61 
62 int main()
63 {
64  //   freopen("in.txt","r",stdin);
65     int i,j;
66     while(~scanf("%d",&n) && n)
67     {
68         for(i=0;i<=n;i++){
69             scanf("%d",&cou[i]);
70             for(j=0;j<cou[i];j++)
71                 scanf("%lf%lf",&nod[i][j].x,&nod[i][j].y);
72         }
73 
74         for(i=1;i<=n;i++){
75             if(cou[i]!=cou[0])continue;
76             if(judge(i,0,1) || judge(i,cou[i]-1,-1))
77                 printf("%d\n",i);
78         }
79         printf("+++++\n");
80     }
81     return 0;
82 }