HDU 2150 Pipe( 判断线段相交水 )


**链接:****传送门 **

题意:

思路:数据量很小,直接暴力所有线段


/*************************************************************************
    > File Name: hdu2150.cpp
    > Author:    WArobot 
    > Blog:      http://www.cnblogs.com/WArobot/ 
    > Created Time: 2017年05月07日 星期日 23时09分58秒
 ************************************************************************/

#include<bits/stdc++.h>
using namespace std;

#define eps 1e-10
struct point{ double x,y; };

bool inter(point a,point b,point c,point d){
	if( min(a.x,b.x) > max(c.x,d.x) ||
		min(a.y,b.y) > max(c.y,d.y) ||
		min(c.x,d.x) > max(a.x,b.x) ||
		min(c.y,d.y) > max(a.y,b.y)	)	return 0;
	double h , i , j , k;
	h = (b.x - a.x)*(c.y - a.y) - (b.y - a.y)*(c.x - a.x);
	i = (b.x - a.x)*(d.y - a.y) - (b.y - a.y)*(d.x - a.x);
	j = (d.x - c.x)*(a.y - c.y) - (d.y - c.y)*(a.x - c.x);
	k = (d.x - c.x)*(b.y - c.y) - (d.y - c.y)*(b.x - c.x);
	return h*i<=eps && j*k<=eps;
}
int main(){
	int N,num[31];
	point pi[31][101];
	while(~scanf("%d",&N)){
		for(int i=0;i<N;i++){
			scanf("%d",&num[i]);
			for(int j=0;j<num[i];j++){
				scanf("%lf%lf",&pi[i][j].x,&pi[i][j].y);
			}
		}
		bool ok = false;
		for(int i=0;i<N;i++){
			for(int j=i+1;j<N;j++){
				for(int k=0;k<num[i]-1;k++){
					for(int s=0;s<num[j]-1;s++){
						if( inter(pi[i][k],pi[i][k+1],pi[j][s],pi[j][s+1])){
							ok = true;	break;
						}
					}
				}
			}
		}
		if(ok)	printf("Yes\n");
		else	printf("No\n");
	}
	return 0;
}
posted @ 2017-05-07 23:32  ojnQ  阅读(249)  评论(0编辑  收藏  举报