Rectangle and Square（判断正方形、矩形）

http://acm.sdut.edu.cn:8080/vjudge/contest/view.action?cid=42#problem/D

改了N多次之后终于A了，一直在改判断正方形和矩形那，判断正方形时算出六条边再排序，若前四条边相等并且与后两条边满足勾股定理，说明是正方形，

判断矩形时，我先对结构体二级排序，这样四个点有确定的顺序，再用点积判断是否有三个角是直角，是的话就是矩形。

  1 #include<stdio.h>

  2 #include<string.h>

  3 #include<algorithm>

  4 using namespace std;

  5 

  6 struct node

  7 {

  8     int x,y;

  9 }point[10];

 10 

 11 int cmp(const struct node a,const struct node b)

 12 {

 13     if(a.x == b.x)

 14         return a.y < b.y;

 15     return a.x < b.x;

 16 }

 17 int dot(const struct node a, const struct node b,const struct node c, const struct node d)

 18 {

 19     int ans = (a.x-b.x)*(c.x-d.x) + (a.y-b.y)*(c.y-d.y);

 20     if(ans == 0) return 1;

 21     return 0;

 22 }

 23 

 24 int dis(const struct node a, const struct node b)

 25 {

 26      return (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y);

 27 }

 28 

 29 int main()

 30 {

 31     while(~scanf("%d %d",&point[1].x,&point[1].y))

 32     {

 33         bool vis[9] = {false};

 34         for(int i = 2; i <= 8; i++)

 35             scanf("%d %d",&point[i].x,&point[i].y);

 36         int flag = 0;

 37         int cnt;

 38         struct node t_point[10];

 39         for(int i = 1; i <= 8; i++)

 40         {

 41             for(int j = i+1; j <= 8; j++)

 42             {

 43                 for(int k = j+1; k <= 8; k++)

 44                 {

 45                     for(int l = k+1; l <= 8; l++)

 46                     {

 47                         int distance[10];

 48                         distance[0] = dis(point[i],point[j]);

 49                         distance[1] = dis(point[i],point[k]);

 50                         distance[2] = dis(point[i],point[l]);

 51                         distance[3] = dis(point[j],point[k]);

 52                         distance[4] = dis(point[j],point[l]);

 53                         distance[5] = dis(point[k],point[l]);

 54                         sort(distance,distance+6);

 55                         if( distance[0] == distance[1] &&

 56                             distance[1] == distance[2] &&

 57                             distance[2] == distance[3] &&

 58                             distance[4] == distance[5] &&

 59                             (distance[0] + distance[1] == distance[5]))

 60                             {

 61                                 flag = 1;

 62                                 vis[i] = true;

 63                                 vis[j] = true;

 64                                 vis[k] = true;

 65                                 vis[l] = true;

 66                                 break;

 67                             }

 68                     }

 69                     if(flag) break;

 70                 }

 71                 if(flag) break;

 72             }

 73             if(flag) break;

 74         }

 75         if(!flag)

 76             printf("NO\n");

 77 

 78         else

 79         {

 80             int tmp1[5],tmp2[5],t1 = 0,t2 = 0;

 81             cnt = 0;

 82             for(int i = 1; i <= 8; i++)

 83             {

 84                 if(!vis[i])

 85                 {

 86                     t_point[cnt++] = point[i];

 87                     tmp2[t2++] = i;

 88                 }

 89                 else tmp1[t1++] = i;

 90             }

 91             sort(t_point,t_point+cnt,cmp);

 92 

 93             if(dot(t_point[0],t_point[1],t_point[0],t_point[2]) &&

 94                dot(t_point[0],t_point[1],t_point[1],t_point[3]) &&

 95                dot(t_point[0],t_point[2],t_point[2],t_point[3]))

 96                {

 97                    printf("YES\n");

 98                    for(int i = 0; i < t1-1; i++)

 99                         printf("%d ",tmp1[i]);

100                     printf("%d\n",tmp1[t1-1]);

101                    for(int i = 0; i < t2-1; i++)

102                         printf("%d ",tmp2[i]);

103                     printf("%d\n",tmp2[t2-1]);

104                }

105             else printf("NO\n");

106 

107         }

108 

109     }

110     return 0;

111 }

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