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[N];int map[N][N];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;}