Anger Begins&&http://acm.nbut.cn:8081/Contest/view/id/29/problem/A.xhtml

题意：给你一些路的首尾坐标，求从A路是否能到达B路

思路：先判断线段是否相交建图，然后求多源多点最短路即可，

AC代码：

#include <iostream>
#include<string.h>
#include<algorithm>
#include<cstdio>
#define CLR(arr,val) memset(arr,val,sizeof(arr))
using namespace std;
#define N 105
#define inf 1<<30
#define Max(x, y)(x > y ? x : y)
#define Min(x, y)(x < y ? x : y)
typedef struct
{
double x1;
double y1;
double x2;
double y2;
}Node;
typedef struct
{
double x;
double y;
}Point;
Node node
;
int map

;
double Multi(Point p1, Point p2, Point p3)//叉乘判断点与线的位置关系
{
return (p1.x - p3.x) * (p2.y - p3.y) - (p2.x - p3.x) * (p1.y - p3.y);
}
bool Isintersect(Point a1, Point a2, Point b1, Point b2)//判断两条线段是否相交（含顶点）
{
if (Min(a1.x, a2.x) <= Max(b1.x, b2.x) &&
Min(a1.y, a2.y) <= Max(b1.y, b2.y) &&
Min(b1.x, b2.x) <= Max(a1.x, a2.x) &&
Min(b1.y, b2.y) <= Max(a1.y, a2.y) &&
Multi(a1, a2, b1) * Multi(a1, a2, b2) <= 0 &&
Multi(b1, b2, a1) * Multi(b1, b2, a2) <= 0
)
return true;//说明两线段之间相交
return false;
}
void in(int &a)
{
char ch;
while((ch=getchar())<'0'||ch>'9');
for( a=0;ch>='0'&&ch<='9';ch=getchar()) a=a*10+ch-'0';

}
bool ok(int a,int b)
{
Point a1,b1,a2,b2;
a1.x=node[a].x1,a1.y=node[a].y1;
b1.x=node[a].x2,b1.y=node[a].y2;
a2.x=node[b].x1,a2.y=node[b].y1;
b2.x=node[b].x2,b2.y=node[b].y2;
if(Isintersect(a1,b1,a2,b2)) return true;
else return false;

}
int main()
{
int n,m;
while(~scanf("%d",&n))
{
for(int i=1;i<=n;++i)
scanf("%lf%lf%lf%lf",&node[i].x1,&node[i].y1,&node[i].x2,&node[i].y2);
for(int i=1;i<=n;++i)
for(int j=1;j<=n;++j)
{
if(ok(i,j)) map[i][j]=map[j][i]=1;
else map[i][j]=map[j][i]=0;
}
in(m);
for(int k=1;k<=n;++k)
for(int i=1;i<=n;++i)
for(int j=1;j<=n;++j)
if(map[k][j]&&map[j][i]) map[k][i]=map[i][k]=1;
for(int i=0;i!=m;++i)
{
int a,b;
in(a),in(b);
if(map[a][b]) puts("YES");
else puts("NO");
}
}
return 0;
}