POJ-2684 Polygonal Line Search 计算几何基础

　　题目链接：http://poj.org/problem?id=2684

　　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 }