跨产品的利用率推断点线段向左或向右，然后你可以2分钟

代码：

#include 
     
      #include 
      
       #include 
       
        using namespace std;const int N = 5005;int n, m, x1, y1, x2, y2;struct Point {    int x, y;    Point() {}    Point(int x, int y) {        this->x = x;        this->y = y;    }};typedef Point Vector;Vector operator - (Vector A, Vector B) {    return Vector(A.x - B.x, A.y - B.y);}struct Seg {    Point a, b;    Seg() {}    Seg(Point a, Point b) {        this->a = a;        this->b = b;    }} seg[N];int ans[N];int Cross(Vector A, Vector B) {return A.x * B.y - A.y * B.x;}void gao(Point p) {    int l = 0, r = n;    while (l < r) {        int mid = (l + r) / 2;        if (Cross(seg[mid].a - p, seg[mid].b - p) < 0) r = mid;        else l = mid + 1;    }    ans[l]++;}int main() {    while (~scanf("%d", &n) && n) {        memset(ans, 0, sizeof(ans));        scanf("%d%d%d%d%d", &m, &x1, &y1, &x2, &y2);        int x, y;        for (int i = 0; i < n; i++) {            scanf("%d%d", &x, &y);            seg[i] = Seg(Point(x, y1), Point(y, y2));        }        for (int i = 0; i < m; i++) {            scanf("%d%d", &x, &y);            gao(Point(x, y));        }        for (int i = 0; i <= n; i++)            printf("%d: %d\n", i, ans[i]);        printf("\n");    }    return 0;}