POJ - 1066 Treasure Hunt

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/lidengdengter/article/details/82049023

题目链接

四堵墙围成的正方形内有n堵墙，求从正方形外进入宝藏处需要破墙次数的最小值。（每次破墙只能从中点破）。

简单想就是从正方形边界找点到宝藏处会与多少条线段(墙)相交，再加上打破边界的这堵墙，求个最小值即可，判断相交用交叉积。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=100;
int n;

struct point{
	double x,y;
	point(){}
	point(double sx,double sy):x(sx),y(sy){	};
	point operator-(const point &w)const{
		return point(x-w.x,y-w.y);
	}
	double operator*(const point &w)const{
		return x*w.x+y*w.y;
	}
	double operator^(const point &w)const{
		return x*w.y-y*w.x;
	}
}p[N],ed;

int work(point a,point b){
	int res=0;
	if(a.x==b.x&&a.y==b.y)
		return 0;
		
	for(int i=1;i<=n;i++){
		double res1=((a-p[i])^(a-b))*((a-p[i+n])^(a-b));
		double res2=((p[i]-a)^(p[i]-p[i+n]))*((p[i]-b)^p[i]-p[i+n]);
		if(res1<=0&&res2<=0)
			res++;
	}
	return res;
}

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%lf%lf%lf%lf",&p[i].x,&p[i].y,&p[i+n].x,&p[i+n].y);
	scanf("%lf%lf",&ed.x,&ed.y);
	
	if(n==0){
		printf("Number of doors = 1\n");
		return 0;
	}
	int ans=1000;
	for(int i=1;i<=2*n;i++)
		ans=min(ans,work(p[i],ed));
	printf("Number of doors = %d\n",ans);
	
	return 0;
}