poj 1066 Treasure Hunt（判断线段相交）

//以下为原blog搬迁过来的内容

【题目大意】：给定一个矩形，和矩形内的墙，矩阵内有一个金矿，问你从矩阵外面进去直到金矿要至少穿越多少墙。

 

【解题思路】：一拿到题就没头没脑的枚举了所有边上的点，然后再直接就判线段相交。1A。结果反过来想，好像这种做法有点问题。看了discuss，有枚举终点的，有旋转的各种做法。以后再一一尝试。不过怎么感觉自己写的都不是正解。

 

【代码】：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>
#include <string>
#include <cctype>
#include <map>
#include <iomanip>
                   
using namespace std;
                   
#define eps 1e-8
#define pi acos(-1.0)
#define inf 1<<30
#define pb push_back
#define lc(x) (x << 1)
#define rc(x) (x << 1 | 1)
#define lowbit(x) (x & (-x))
#define ll long long

struct Point
{
    double x, y;
    Point() {}
    Point(double a, double b)
    {
        x = a, y = b;
    }
}point[1000];

struct Line
{
    Point a, b;
    Line() {}
    Line(Point x, Point y)
    {
        a = x, b = y;
    }
}line[1000];

inline int sig(double k)
{
    return k < -eps ? -1 : k > eps;
}

inline double det(double x1, double y1, double x2, double y2)
{
    return x1 * y2 - x2 * y1;
}
inline double xmult(Point o, Point a, Point b)
{
    return det(a.x - o.x, a.y - o.y, b.x - o.x, b.y - o.y);
}

inline int intersect1(Point a, Point b, Point c, Point d) {
    double s1, s2, s3, s4;
    int d1 = sig(s1 = xmult(a, b, c));
    int d2 = sig(s2 = xmult(a, b, d));
    int d3 = sig(s3 = xmult(c, d, a));
    int d4 = sig(s4 = xmult(c, d, b));
    if ((d1^d2) == -2 && (d3^d4) == -2)
    {
        return 1;
    }
    return 0;
}

inline int intersect(Line u, Line v)
{
    return intersect1(u.a, u.b, v.a, v.b);
}

int main()
{
    int n;
    double p,q;
    scanf("%d",&n);
    for (int i=1; i<=n; i++)
    {
        scanf("%lf%lf",&p,&q);
        point[i*2-1]=Point(p,q);
        scanf("%lf%lf",&p,&q);
        point[i*2]=Point(p,q);
        line[i]=Line(point[2*i-1],point[2*i]);
    }
    scanf("%lf%lf",&p,&q);
    Point goal;
    goal=Point(p,q);
    int cnt,minn;
    minn=1<<30;
    for (int i=0; i<=2*n; i++)
    {
        cnt=1;
        for (int j=1; j<=n; j++)
        {
            Line tmp;
            tmp=Line(point[i],goal);
            int k;
            k=intersect(tmp,line[j]);
            if (k==1) cnt++;
        }
        if (cnt<=minn) minn=cnt;
    }
    printf("Number of doors = ");
    printf("%d\n",minn);
    return 0;
}