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 }
View Code

 

posted on 2013-11-10 20:38  straw_berry  阅读(412)  评论(0编辑  收藏  举报