POJ2526+简单几何

题意：给定的这些点是否有一个对称中心。

PS：我写得有点啰嗦。。

就是把小的x和大的x进行匹配。

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<math.h>
  4 using namespace std;
  5 
  6 const double eps = 1e-8;
  7 const int maxn = 10005;
  8 
  9 struct Point{
 10     double x,y;
 11 };
 12 Point pnt1[ maxn ],pnt2[ maxn ];
 13 Point cc;
 14 
 15 int cmp1( Point a,Point b ){
 16     return a.x<b.x;
 17 }
 18 
 19 int cmp2( Point a,Point b ){
 20     return a.y<b.y;
 21 }
 22 
 23 int cmp3( Point a,Point b ){
 24     return a.x>b.x;
 25 }
 26 
 27 bool NotOnePoint( Point a,Point b ){
 28     if( fabs(a.x-b.x)<=eps&&fabs(a.y-b.y)<=eps ) return false;
 29     else return true;
 30 }
 31 
 32 bool OK( Point a,Point b ){
 33     if( fabs((a.x+b.x)-(2.0*cc.x))<=eps && fabs((a.y+b.y)-(2.0*cc.y))<=eps ) return true;
 34     else return false;
 35 }
 36 
 37 bool Notcc( Point a ){
 38     if( fabs(a.x-cc.x)<=eps&&fabs(a.y-cc.y)<=eps ) return true;
 39     else return false;
 40 }
 41 
 42 int main(){
 43     int n;
 44     int T;
 45     //freopen("in.txt","r",stdin);
 46     scanf("%d",&T);
 47     while( T-- ){
 48         scanf("%d",&n);
 49         for( int i=1;i<=n;i++ ){
 50             scanf("%lf%lf",&pnt1[i].x,&pnt1[i].y);
 51             pnt2[ i ] = pnt1[ i ];
 52         }
 53         if( n==2 ){
 54             puts("yes");
 55             continue;
 56         }
 57         sort( pnt1+1,pnt1+1+n,cmp1 );
 58         cc.x = (pnt1[1].x+pnt1[n].x)/2.0;
 59         sort( pnt2+1,pnt2+1+n,cmp2 );
 60         cc.y = (pnt2[1].y+pnt2[n].y)/2.0;
 61         sort( pnt2+1,pnt2+1+n,cmp3 );
 62 
 63         int cnt = 0;
 64         if( n%2==1 ){
 65             for( int i=1;i<=n;i++ ){
 66                 if( pnt1[i].x==cc.x&&pnt1[i].y==cc.y ){
 67                     cnt++;
 68                 }
 69             }
 70         }
 71         if( (n-cnt)%2==1 ){
 72             puts("no");
 73             continue;
 74         }
 75         if( n==cnt ){
 76             puts("yes");
 77             continue;
 78         }
 79         //printf("cc:x = %lf,y = %lf\n",cc.x,cc.y);
 80         /*
 81         for( int i=1;i<=n;i++ ){
 82             printf("x = %lf \n",pnt1[i].x);
 83         }
 84         for( int i=1;i<=n;i++ ){
 85             printf("x = %lf \n",pnt2[i].x);
 86         }
 87         */
 88         int tt = 0;
 89         int N = n - cnt;//N%2=0
 90         for( int i=1;i<=(n/2)&&tt<(N/2);i++ ){
 91             tt++;
 92             //printf(" i =%d ",i);
 93             //printf("pnt1:x = %lf y = %lf\n",pnt1[i].x,pnt1[i].y);
 94             bool find = false;
 95             for( int j=1;j<=(n/2);j++ ){
 96                 //printf(" j = %d \n",j);
 97                 //printf("pnt2:x = %lf y = %lf\n",pnt2[j].x,pnt2[j].y);
 98                 if( (pnt1[i].x+pnt2[j].x)<2.0*cc.x ) break;
 99                 if( /*Notcc(pnt1[i])==true&&*/NotOnePoint(pnt1[i],pnt2[j])==true&&OK(pnt1[i],pnt2[j])==true ){
100                     find = true;
101                     cnt += 2;
102                     break;
103                 }
104             }
105             //if( find==true ) printf("true\n");
106             //else printf("false\n");
107             if( find==false ) break;
108         }
109         if( cnt==n ){
110             puts("yes");
111             continue;
112         }
113         puts("no");
114     }
115     return 0;
116 }
View Code

 

posted @ 2013-08-23 23:00  xxx0624  阅读(185)  评论(0)  编辑  收藏  举报