POJ 1066 Treasure Hunt （线段相交）

#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<vector>
#include<string>
#include<cstdio>
#include<cstring>
#include<iomanip>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define eps 1E-8
/*注意可能会有输出-0.000*/
#define Sgn(x) (x<-eps? -1 :x<eps? 0:1)//x为两个浮点数差的比较,注意返回整型
#define Cvs(x) (x > 0.0 ? x+eps : x-eps)//浮点数转化
#define zero(x) (((x)>0?(x):-(x))<eps)//判断是否等于0
#define mul(a,b) (a<<b)
#define dir(a,b) (a>>b)
typedef long long ll;
typedef unsigned long long ull;
const int Inf=1<<28;
const ll INF=1ll<<60;
const double Pi=acos(-1.0);
const int Mod=1e9+7;
const int Max=500;
struct point
{
    double x,y;
};
point poi[Max],enn;
double xmult(point p1,point p2,point p0)//计算 cross product (p1-p0) x (p2-p0)
{
    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}

int isIntersected(point p1,point p2,point l1,point l2)//线段相交（不包括端点重合）
{
    return (max(p1.x,p2.x)>=min(l1.x,l2.x)) &&
           (max(p1.y,p2.y)>=min(l1.y,l2.y)) &&
           (max(l1.x,l2.x)>=min(p1.x,p2.x)) &&
           (max(l1.y,l2.y)>=min(p1.y,p2.y)) &&
           (xmult(l1,p2,p1)*xmult(p2,l2,p1)>0) &&
           (xmult(p1,l2,l1)*xmult(l2,p2,l1)>0) ;
}

int Solve(int n)
{
    int minx=Inf;
    for(int i=0; i<n; ++i)//枚举每个端点
    {
        int res=1;
        for(int j=0; j<n-4; j+=2)//每堵墙找是否有交点
        {
            if(isIntersected(poi[i],enn,poi[j],poi[j+1]))
                res++;
        }
        minx=min(minx,res);
    }
    if(!n)
        return 1;
    return minx;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        n<<=1;
        for(int i=0; i<n; ++i)
        {
            scanf("%lf %lf",&poi[i].x,&poi[i].y);
        }
        scanf("%lf %lf",&enn.x,&enn.y);
        poi[n].x=0,poi[n++].y=0;
        poi[n].x=0,poi[n++].y=100;
        poi[n].x=100,poi[n++].y=0;
        poi[n].x=100,poi[n++].y=100;//添加外墙四个点
        printf("Number of doors = %d\n",Solve(n));
    }
    return 0;
}