POJ 2318 (叉积) TOYS

题意：

有一个长方形，里面从左到右有n条线段，将矩形分成n+1个格子，编号从左到右为0~n。

端点分别在矩形的上下两条边上，这n条线段互不相交。

现在已知m个点，统计每个格子中点的个数。

分析：

用叉积判断点与线段的相对位置，对于每个点二分查找所在的格子。

#include <cstdio>
#include <cmath>
#include <cstring>

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, kase = 0;
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]);
        for(int i = 0; i < m; ++i)
        {
            Point P; P = read_point();
            int pos = binary_search(P);
            ans[pos]++;
        }

        if(kase++) puts("");
        for(int i = 0; i <= n; ++i) printf("%d: %d\n", i, ans[i]);
    }

    return 0;
}