poj2398 计算几何

poj2318的升级版，只是中间的n个隔板未排序，输出的是装有i个玩具的区域的个数，其他完全一样

poj2318的解题报告见http://blog.youkuaiyun.com/non_cease/article/details/7754382

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

struct Point {
    int x, y;
};

struct Line {
   Point a, b;
} line[5005];

int cnt[5005], ans[5005];

int Min(int a, int b) {
    return a > b ? b : a;
}

int Max(int a, int b) {
    return a > b ? a : b;
}

bool cmp(const Line& l1, const Line& l2) {
    if (Min(l1.a.x, l1.b.x) == Min(l2.a.x, l1.b.x))
        return Max(l1.a.x, l1.b.x) < Max(l2.a.x, l1.b.x);
    return Min(l1.a.x, l1.b.x) < Min(l2.a.x, l1.b.x);
}

int Multi(Point p1, Point p2, Point p0) {
    return (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);
}

void BSearch(Point a, int n) {
    int l, r, mid;

    l = 0; r = n-1;
    while (l < r) {
        mid = (l + r) >> 1;
        if (Multi(a, line[mid].a, line[mid].b) > 0) l = mid + 1;
        else r = mid;
    }
    if (Multi(a, line[l].a, line[l].b) < 0) cnt[l]++;
    else cnt[l+1]++;
}

int main()
{
    int n, m, x1, y1, x2, y2;
    int i, t1, t2;
    Point a;

    while (scanf ("%d", &n) && n) {
        scanf ("%d%d%d%d%d", &m, &x1, &y1, &x2, &y2);
        for (i = 0; i < n; i++) {
            scanf ("%d%d", &t1, &t2);
            line[i].a.x = t1;
            line[i].a.y = y1;
            line[i].b.x = t2;
            line[i].b.y = y2;
        }
        sort(line, line+n, cmp);
        memset(cnt, 0, sizeof (cnt));
        memset(ans, 0, sizeof (ans));
        for (i = 0; i < m; i++) {
            scanf ("%d%d", &a.x, &a.y);
            BSearch(a, n);
        }
        for (i = 0; i <= n; i++) ans[cnt[i]]++;
        printf ("Box\n");
        for (i = 1; i <= m; i++)
            if (ans[i] != 0) {
                printf ("%d: %d\n", i, ans[i]);
                m -= i * cnt[i];
            }
    }
    return 0;
}