POJ 1066 Treasure Hunt 线段交

来源：互联网发布：php魔方加密在线解密编辑：程序博客网时间：2024/05/16 04:47

要到达目标点。

必然会经过一系列的线段

在这些线段上，在哪里开门都是一样的。

所以只需枚举边上的点到目标的点连接的线段中间交多少线段，求最小即可

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cstdlib>#include <cmath>#include <map>#include <sstream>#include <queue>#include <vector>#define MAXN 100005#define MAXM 211111#define eps 1e-8#define INF 500000001using namespace std;inline int dblcmp(double d){    if(fabs(d) < eps) return 0;    return d > eps ? 1 : -1;}struct point{    double x, y;    point(){}    point(double _x, double _y): x(_x), y(_y) {}    void input()    {        scanf("%lf%lf", &x, &y);    }    bool operator ==(point a)const    {        return dblcmp(a.x - x) == 0 && dblcmp(a.y - y) == 0;    }    point sub(point p)    {        return point(x - p.x, y - p.y);    }    double dot(point p)    {        return x * p.x + y * p.y;    }    double det(point p)    {        return x * p.y - y * p.x;    }    double distance(point p)    {        return hypot(x - p.x, y - p.y);    }}p[111], des;struct line{    point a, b;    line(){}    line(point _a, point _b){ a = _a; b = _b;}    void input()    {        a.input();        b.input();    }    int segcrossseg(line v)    {        int d1 = dblcmp(b.sub(a).det(v.a.sub(a)));        int d2 = dblcmp(b.sub(a).det(v.b.sub(a)));        int d3 = dblcmp(v.b.sub(v.a).det(a.sub(v.a)));        int d4 = dblcmp(v.b.sub(v.a).det(b.sub(v.a)));        if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2)return 2;        return (d1 == 0 && dblcmp(v.a.sub(a).dot(v.a.sub(b))) <= 0||                d2 == 0 && dblcmp(v.b.sub(a).dot(v.b.sub(b))) <= 0||                d3 == 0 && dblcmp(a.sub(v.a).dot(a.sub(v.b))) <= 0||                d4 == 0 && dblcmp(b.sub(v.a).dot(b.sub(v.b))) <= 0);    }}seg[111];int n, m, t;int main(){    p[n++] = point(0, 0);    p[n++] = point(0, 100);    p[n++] = point(100, 0);    p[n++] = point(100, 100);    scanf("%d", &t);    for(int i = 0; i < t; i++)    {        seg[i].input();        p[n++] = seg[i].a;        p[n++] = seg[i].b;    }    seg[t++] = line(p[0], p[1]);    seg[t++] = line(p[0], p[2]);    seg[t++] = line(p[1], p[3]);    seg[t++] = line(p[2], p[3]);    des.input();    int ans = INF;    for(int i = 0; i < n; i++)    {        line tmp = line(p[i], des);        int cnt = 0;        for(int j = 0; j < t; j++)        {            if(tmp.segcrossseg(seg[j])) cnt++;        }        if(cnt < ans) ans = cnt;    }    printf("Number of doors = %d\n", ans - 1);    return 0;}