POJ 2398 Toy Storage

这道题和POJ 2318几乎是一样的。

区别就是输入中坐标不给排序了，=_=||

输出变成了，有多少个区域中有t个点。

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

struct Point
{
    int x, y;
    Point(int x=0, int y=0):x(x), y(y) {}
};
typedef Point Vector;

Point read_point()
{
    int x, y;
    scanf("%d%d", &x, &y);
    return Point(x, y);
}

Point operator + (const Point& A, const Point& B)
{ return Point(A.x+B.x, A.y+B.y); }

Point operator - (const Point& A, const Point& B)
{ return Point(A.x-B.x, A.y-B.y); }

int Cross(const Point& A, const Point& B)
{ return A.x*B.y - A.y*B.x; }

const int maxn = 5000 + 10;
int up[maxn], down[maxn], ans[maxn];
int n, m;
Point A0, B0;

int binary_search(const Point& P)
{
    int L = 0, R = n;
    while(L < R)
    {
        int mid = L + (R - L + 1) / 2;
        Point A(down[mid], B0.y), B(up[mid], A0.y);
        Vector v1 = B - A;
        Vector v2 = P - A;
        if(Cross(v1, v2) < 0) L = mid;
        else R = mid - 1;
    }
    return L;
}

int main()
{
    //freopen("in.txt", "r", stdin);

    while(scanf("%d", &n) == 1 && n)
    {
        memset(ans, 0, sizeof(ans));

        scanf("%d", &m); A0 = read_point(); B0 = read_point();
        for(int i = 1; i <= n; ++i) scanf("%d%d", &up[i], &down[i]);
        sort(up + 1, up + 1 + n); sort(down + 1, down + 1 + n);
        for(int i = 0; i < m; ++i)
        {
            Point P; P = read_point();
            int pos = binary_search(P);
            ans[pos]++;
        }
        sort(ans, ans + n + 1);

        puts("Box");
        int i;
        for(i = 0; i <= n; ++i) if(ans[i]) break;
        while(i <= n)
        {
            int cnt = 1;
            int num = ans[i];
            while(i <= n && ans[i+1] == ans[i]) { i++; cnt++; }
            i++;
            printf("%d: %d\n", num, cnt);
        }
    }

    return 0;
}