Poj-1066 (判断线段与线段是否相交)

题目

分析

程序

#include <cstdio>
int n,ans=100;
struct node{double x,y;} U[50],V[50],P;
double cheng(node A,node B){return (A.x*B.y)-(B.x*A.y);}
node vct(node A,node B){return (node){B.x-A.x,B.y-A.y};}
void zzk(node X,node Y){
int ret=0;
double k1,k2,k3,k4;
for (int i=1; i<=n; i++){
k1=cheng(vct(X,Y),vct(X,U[i]));
k2=cheng(vct(X,Y),vct(X,V[i]));
k3=cheng(vct(U[i],V[i]),vct(U[i],X));
k4=cheng(vct(U[i],V[i]),vct(U[i],Y));
if (k1*k2<=0 && k3*k4<=0) ret++;
}
if (ret<ans) ans=ret;
}

int main(){

scanf("%d",&n);
if (n==0) ans=1;    //要是没有线段，那么只用炸最外面一个墙即可
for (int i=1; i<=n; i++) scanf("%lf%lf%lf%lf",&U[i].x,&U[i].y,&V[i].x,&V[i].y);
scanf("%lf%lf",&P.x,&P.y);
for (int i=1; i<=n; i++){
zzk(P,U[i]);
zzk(P,V[i]);
}
printf("Number of doors = %d",ans);
}

提示