看上去像是一个最大权闭合子图裸题但是数据太大

我们可以先把守卫的视野转换到第二象限(每个守卫可以看到横坐标比他小 纵坐标比他大的宝物) 然后按X从小到大 再按Y从大到小排

这样我们就可以按SORT序遍历守卫 然后贪心地把每个守卫的流量流给离他最近的Y最小的宝物 易证这样是最优的

#include<bits/stdc++.h>
#define ll long long
using namespace std;
inline void read(int &x) {
        char c;
        int f = 1;
        while (!((c = getchar()) >= '0' && c <= '9'))
                if (c == '-')
                        f = -1;
        x = c - '0';
        while ((c = getchar()) >= '0' && c <= '9')
                (x *= 10) += c - '0';
        if (f == -1)
                x = -x;
}
const int maxn = 810000;
const double eps = 1e-9;
int n, m, N, w, h;
struct node {
        double x, y;
        int i, c;
} X, Y, a[maxn];
inline bool operator <(const node x, const node y) {
        return x.y - y.y > eps;
}
inline bool cmp(const node x, const node y) {//按横坐标从小到大，纵坐标从大到小排序
        if (fabs(x.x - y.x) < eps)
                return x.y - y.y < -eps;
        return x.x - y.x < -eps;
}
multiset<node>S;
multiset<node>::iterator it, it2;
node generate(int x, int y) { //视野转换为第二象限
        double A = ((double)x / w - (double)y / h) / 2.0;
        double B = A - (double)x / w;
        return (node) {
                -A, -B, 0, 0
        };
}
ll re;
int main() {
        read(n);
        read(m);
        N = n + m;
        read(w);
        read(h);
        //X=(node){w,-h};
        //Y=(node){-w,-h};
        for (int i = 1; i <= n; i++) {
                int x, y, c;
                read(x);
                read(y);
                read(c);
                a[i] = generate(x, y);
                a[i].i = i;
                a[i].c = c;
                re += c;
        }
        for (int i = 1; i <= m; i++) {
                int x, y, c;
                read(x);
                read(y);
                read(c);
                a[n + i] = generate(x, y);
                a[n + i].i = n + i;
                a[n + i].c = c;
        }
        sort(a + 1, a + N + 1, cmp);
        for (int i = 1; i <= N; i++) {
                if (a[i].i <= n)
                        S.insert(a[i]);
                else {
                        it = S.lower_bound(a[i]);
                        while (it != S.end() && a[i].c) {
                                it2 = it;
                                it2++; //指向下一个
                                if (a[i].c < (*it).c) {
                                        node tmp = (*it);
                                        tmp.c -= a[i].c;
                                        re -= a[i].c;
                                        S.erase(it);
                                        S.insert(tmp);
                                        break;
                                }
                                a[i].c -= (*it).c;
                                re -= (*it).c;
                                S.erase(it);
                                it = it2;
                        }
                }
        }
        printf("%lld\n", re);
        return 0;
}