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。

 #include<stdio.h>
 #include<algorithm>
 #include<math.h>
 using namespace std;

 const double eps = 1e-;
 const int maxn = ;

 struct Point{
     double x,y;
 };
 Point pnt1[ maxn ],pnt2[ maxn ];
 Point cc;

 int cmp1( Point a,Point b ){
     return a.x<b.x;
 }

 int cmp2( Point a,Point b ){
     return a.y<b.y;
 }

 int cmp3( Point a,Point b ){
     return a.x>b.x;
 }

 bool NotOnePoint( Point a,Point b ){
     if( fabs(a.x-b.x)<=eps&&fabs(a.y-b.y)<=eps ) return false;
     else return true;
 }

 bool OK( Point a,Point b ){
     if( fabs((a.x+b.x)-(2.0*cc.x))<=eps && fabs((a.y+b.y)-(2.0*cc.y))<=eps ) return true;
     else return false;
 }

 bool Notcc( Point a ){
     if( fabs(a.x-cc.x)<=eps&&fabs(a.y-cc.y)<=eps ) return true;
     else return false;
 }

 int main(){
     int n;
     int T;
     //freopen("in.txt","r",stdin);
     scanf("%d",&T);
     while( T-- ){
         scanf("%d",&n);
         for( int i=;i<=n;i++ ){
             scanf("%lf%lf",&pnt1[i].x,&pnt1[i].y);
             pnt2[ i ] = pnt1[ i ];
         }
         if( n== ){
             puts("yes");
             continue;
         }
         sort( pnt1+,pnt1++n,cmp1 );
         cc.x = (pnt1[].x+pnt1[n].x)/2.0;
         sort( pnt2+,pnt2++n,cmp2 );
         cc.y = (pnt2[].y+pnt2[n].y)/2.0;
         sort( pnt2+,pnt2++n,cmp3 );

         int cnt = ;
         if( n%== ){
             for( int i=;i<=n;i++ ){
                 if( pnt1[i].x==cc.x&&pnt1[i].y==cc.y ){
                     cnt++;
                 }
             }
         }
         if( (n-cnt)%== ){
             puts("no");
             continue;
         }
         if( n==cnt ){
             puts("yes");
             continue;
         }
         //printf("cc:x = %lf,y = %lf\n",cc.x,cc.y);
         /*
         for( int i=1;i<=n;i++ ){
             printf("x = %lf \n",pnt1[i].x);
         }
         for( int i=1;i<=n;i++ ){
             printf("x = %lf \n",pnt2[i].x);
         }
         */
         int tt = ;
         int N = n - cnt;//N%2=0
         for( int i=;i<=(n/)&&tt<(N/);i++ ){
             tt++;
             //printf(" i =%d ",i);
             //printf("pnt1:x = %lf y = %lf\n",pnt1[i].x,pnt1[i].y);
             bool find = false;
             for( int j=;j<=(n/);j++ ){
                 //printf(" j = %d \n",j);
                 //printf("pnt2:x = %lf y = %lf\n",pnt2[j].x,pnt2[j].y);
                 if( (pnt1[i].x+pnt2[j].x)<2.0*cc.x ) break;
                 if( /*Notcc(pnt1[i])==true&&*/NotOnePoint(pnt1[i],pnt2[j])==true&&OK(pnt1[i],pnt2[j])==true ){
                     find = true;
                     cnt += ;
                     break;
                 }
             }
             //if( find==true ) printf("true\n");
             //else printf("false\n");
             if( find==false ) break;
         }
         if( cnt==n ){
             puts("yes");
             continue;
         }
         puts("no");
     }
     return ;
 }