POJ1066：Treasure Hunt——题解

http://poj.org/problem?id=1066

题目大意：给一个由墙围成的正方形，里面有若干墙，每次破墙只能从（当前看到的）墙的中点破，求最少破多少墙才能看到宝藏。

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显然枚举起点然后画一条以终点为端点的线，求线和多少墙相交即可。

但是我不会证明：这里有证明：https://www.cnblogs.com/anrainie/p/4049844.html

然后有该证明我们得到只要起点为这些线的端点即可。

#include<cstdio>
#include<queue>
#include<cctype>
#include<cstring>
#include<vector>
#include<cmath>
#include<algorithm>
using namespace std;
typedef double dl;
const int INF=2147483647;
const int N=31;
struct point{//既是向量又是点
    dl x;
    dl y;
}p[2*N],ed;
int n;
inline point getmag(point a,point b){
    point s;
    s.x=b.x-a.x;s.y=b.y-a.y;
    return s;
}
inline dl multiX(point a,point b){
    return a.x*b.y-b.x*a.y;
}
inline int intersect(point a,point b){
    int ans=0;
    if(a.x==b.x&&a.y==b.y)return 0;
    for(int i=1;i<=n;i++){
    dl d1=multiX(getmag(a,p[i]),getmag(a,b))*multiX(getmag(a,p[i+n]),getmag(a,b));
    dl d2=multiX(getmag(p[i],a),getmag(p[i],p[i+n]))*multiX(getmag(p[i],b),getmag(p[i],p[i+n]));
    if(d1<=0&&d2<=0)ans++;
    }
    return ans;
}
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
");
    return 0;
    }
    int ans=INF;
    for(int i=1;i<=2*n;i++){
    ans=min(ans,intersect(p[i],ed));
    }
    printf("Number of doors = %d
",ans);
    return 0;
}