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。。。。我也就写写这么挫的东东。。。

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-8;
const int maxn = 10005;

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=1;i<=n;i++ ){
			scanf("%lf%lf",&pnt1[i].x,&pnt1[i].y);
			pnt2[ i ] = pnt1[ i ];
		}
		if( n==2 ){
			puts("yes");
			continue;
		}
		sort( pnt1+1,pnt1+1+n,cmp1 );
		cc.x = (pnt1[1].x+pnt1[n].x)/2.0;
		sort( pnt2+1,pnt2+1+n,cmp2 );
		cc.y = (pnt2[1].y+pnt2[n].y)/2.0;
		sort( pnt2+1,pnt2+1+n,cmp3 );

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