POJ 1066 Treasure Hunt

题意：在金字塔内有一个宝藏p(x,y)。现在要取出这个宝藏，在金字塔内有许多面墙，为了进入宝藏所在的位置必须把墙炸开，炸墙只能炸每个房间墙的中点，求将宝藏运出城堡所需要的最小炸墙数。

分析：虽然炸墙只能炸中点，但是因为每个房间里面是空的，而且关心的是最少要炸几面墙，金字塔范围因为只有100*100，所以直接枚举边上的每个点与宝藏的连线，与墙的交点个数最少的直线，就是炸墙的路线。

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<cstdio>
#include<vector>
#include <iomanip>
#include<utility>
#include<cctype>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define inf 0x3f3f3f3f
#define Clear(x) memset(x,0,sizeof(x))
#define fup(i,a,b) for(int i=a;i<b;i++)
#define rfup(i,a,b) for(int i=a;i<=b;i++)
#define fdn(i,a,b) for(int i=a;i>b;i--)
#define rfdn(i,a,b) for(int i=a;i>=b;i--)
typedef long long ll;
using namespace std;
const double pi=acos(-1.0);
const int maxn = 35;
const double eps = 1e-8;
int n;

struct Point{
    double x,y;
    Point(){}
    Point(double _x,double _y){x=_x;y=_y;}
}trea;

struct Line{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e){s=_s;e=_e;}
}line[maxn];

int sgn(double x)
{
    if(fabs(x)<eps) return 0;
    else return x<0?-1:1;
}

double cross(Point a,Point b,Point c)
{
    a.x-=c.x;a.y-=c.y;
    b.x-=c.x;b.y-=c.y;
    return a.x*b.y-b.x*a.y;
}

bool intersect(Line l1,Line l2)
{
    int d1=sgn(cross(l1.s,l2.s,l2.e));
    int d2=sgn(cross(l1.e,l2.s,l2.e));
    int d3=sgn(cross(l2.e,l1.s,l1.e));
    int d4=sgn(cross(l2.s,l1.s,l1.e));
    if(d1*d2<0&&d3*d4<0)
        return true;
    return false;
}

int slove(Line l1)
{
    int ret=0;
    for(int i=1;i<=n;i++)
        if(intersect(l1,line[i]))
            ret++;
    return ret;
}

int main()
{
    double x1,y1,x2,y2;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
        line[i]=Line(Point(x1,y1),Point(x2,y2));
    }
    scanf("%lf%lf",&trea.x,&trea.y);
    Line t;
    int ans=inf;
    for(int i=1;i<=100;i++)
    {
        t.s=trea;
        t.e.x=i,t.e.y=0,ans=min(ans,slove(t));
        t.e.x=i,t.e.y=100,ans=min(ans,slove(t));
        t.e.x=0,t.e.y=i,ans=min(ans,slove(t));
        t.e.x=100,t.e.y=i,ans=min(ans,slove(t));
    }
    printf("Number of doors = %d\n",ans+1);
    return 0;
}