poj 1066 Treasure Hunt

枚举边上相邻点的中点，和终点连线，计算其和其他线段的交点。

值得回味的地方是kuailezhish帮我想出的排序的方案。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<map>
#include<stack>
#include<queue>
#include<algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define MAX(a,b) ((a)>(b)?(a):(b))
#define MIN(a,b) ((a)<(b)?(a):(b))
#define INF 0x3f3f3f3f
using namespace std;

struct point {
    double x, y;
    void setPoint(double xx, double yy) {
        x = xx;
        y = yy;
    }
};

struct line {
    point s, e;
};

int n;
line num[109];
point ed, a[509];
int acnt;
double eps = 1e-8;

int getval(point a) {
    int ret = 0;
    if(a.x <= eps) {
        ret = a.y;
    } else if(fabs(a.x - 100) <= eps) {
        ret = 101 + a.y;
    } else if(a.y <= eps) {
        ret = 202 + a.x;
    } else {
        ret = 303 + a.x;
    }
    return ret;
}

bool cmp1(point a, point b) {
    int ta = getval(a);
    int tb = getval(b);
    return ta < tb;
}

bool dy(double a, double b) {
    return fabs(a - b) < eps;
}

double mp(point a, point b, point c) {
    return ((b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y));
}

int iscross(line a, line b) {
    if(mp(a.s, a.e, b.s) * mp(a.s, a.e, b.e) < -eps && mp(b.s, b.e, a.s) * mp(b.s, b.e, a.e) < -eps) {
        return 1;
    }
    return 0;
}

int cntline(line t) {
    int ret = 1;
    for(int i = 0; i < n; i++) {
        ret += iscross(t, num[i]);
    }
    return ret;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("F:/ACMData.txt","r",stdin);
#endif
    scanf("%d", &n);
    acnt = 0;
    for(int i = 0; i < n; i++) {
        scanf("%lf%lf%lf%lf", &num[i].s.x, &num[i].s.y, &num[i].e.x, &num[i].e.y);
        a[acnt++] = num[i].s;
        a[acnt++] = num[i].e;
    }
    scanf("%lf%lf", &ed.x, &ed.y);
    a[acnt++].setPoint(0, 0);
    a[acnt++].setPoint(0, 100);
    a[acnt++].setPoint(100, 0);
    a[acnt++].setPoint(100, 100);
    sort(a, a + acnt, cmp1);
    int ret = INF;
    for(int i = 0; i < acnt - 1; i++) {
        if(dy(a[i].x, a[i + 1].x) && dy(a[i].y, a[i + 1].y)) {
            continue;
        }
        if(dy(a[i].x, a[i + 1].x) || dy(a[i].y, a[i + 1].y)) {
            line p;
            p.s = ed;
            p.e.setPoint(a[i].x / 2 + a[i + 1].x / 2, a[i].y / 2 + a[i + 1].y / 2);
            int tmp = cntline(p);
            ret = MIN(ret, tmp);
        }
    }
    printf("Number of doors = %d\n", ret);
    return 0;
}