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传送门
一个点选了就必须选若干个点，最大化点权之和，显然最大权闭合子图问题。
一个点向它范围内所有点连边，直接跑最大权闭合子图即可。
参考代码：

#include <cstring>
#include <cstdio>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
template < class T > inline T min(T a, T b) { return a < b ? a : b; }
template < class T > inline void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while ('0' > c || c > '9') f |= c == '-', c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
    s = f ? -s : s;
}
 
const int _ = 502, __ = 3e5 + 5, INF = 2147483647;
 
int tot = 1, head[_]; struct Edge { int v, w, nxt; } edge[__ << 1];
inline void Add_edge(int u, int v, int w) { edge[++tot] = (Edge) { v, w, head[u] }, head[u] = tot; }
inline void Link(int u, int v, int w) { Add_edge(u, v, w), Add_edge(v, u, 0); }
 
int n, X[_], Y[_], R[_], S[_];
int s, t, dep[_], cur[_], hd, tl, Q[_];
 
inline int bfs() {
    memset(dep, 0, sizeof (int) * (t - s + 1));
    hd = tl = 0, dep[Q[++tl] = s] = 1;
    while (hd < tl) {
        int u = Q[++hd];
        for (rg int i = head[u]; i; i = edge[i].nxt) {
            int v = edge[i].v, w = edge[i].w;
            if (dep[v] == 0 && w) dep[v] = dep[u] + 1, Q[++tl] = v;
        }
    }
    return dep[t] > 0;
}
 
inline int dfs(int u, int flow) {
    if (u == t) return flow;
    for (rg int& i = cur[u]; i; i = edge[i].nxt) {
        int v = edge[i].v, w = edge[i].w;
        if (dep[v] == dep[u] + 1 && w) {
            int res = dfs(v, min(flow, w));
            if (res) { edge[i].w -= res, edge[i ^ 1].w += res; return res; }
        }
    }
    return 0;
}
 
inline int Dinic() {
    int res = 0;
    while (bfs()) {
        for (rg int i = s; i <= t; ++i) cur[i] = head[i];
        while (int d = dfs(s, INF)) res += d;
    }
    return res;
}
 
int main() {
#ifndef ONLINE_JUDGE
    file("cpp");
#endif
    read(n), s = 0, t = n + 1;
    int ans = 0;
    for (rg int i = 1; i <= n; ++i) {
        read(X[i]), read(Y[i]), read(R[i]), read(S[i]);
        if (S[i] > 0) Link(s, i, S[i]), ans += S[i];
        else if (S[i] < 0) Link(i, t, -S[i]);
    }
    for (rg int i = 1; i <= n; ++i)
        for (rg int j = 1; j <= n; ++j)
            if (i != j && (X[i] - X[j]) * (X[i] - X[j]) + (Y[i] - Y[j]) * (Y[i] - Y[j]) <= R[i] * R[i]) Link(i, j, INF);
    printf("%d\n", ans - Dinic());
    return 0;
}

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