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题目描述

简要题意：给定一个二维平面，平面上有 $n$ 个白点，第 $i$ 个白点的坐标为 $(x_i,y_i)$，其能传输的范围为 $r_i$，收益为 $s_i$，第 $i$ 个点能传输到第 $j$ 个点，当且仅当两点之间的欧几里得距离小于等于 $r_i$，现在要将某些点涂黑，如果染黑每个点则必须将其能到达的点都染黑，染黑一个点能得到它的收益，求最大收益

$n\le 100$

Solution

注意到原图是有向图，且一个强连通分量必须同时选，我们缩点后得到一个 $DAG$，然后我们传递闭包，能够发现，现在我们选一个点需要选它连出去的所有点

容易发现这个东西非常像是最大权闭合子图的模型，我们将点分成正权点和负权点，起点连正权点，终点连负权点，然后正权点连它需要的负权点，注意两个正权点之间的依赖关系不会影响

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#define maxn 110
#define ll long long
#define INF 1000000000
using namespace std;

struct Edge {
    int to, next, w;
} e[1000000]; int c1, head[maxn];
inline void add_edge(int u, int v, int w) {
    e[c1].to = v; e[c1].w = w;
    e[c1].next = head[u]; head[u] = c1++;
}

inline void Add_edge(int u, int v, int w) {
    add_edge(u, v, w); add_edge(v, u, 0);
}

int dep[maxn], s, t; 
bool bfs() {
    queue<int> Q; Q.push(s); 
    fill(dep, dep + maxn, 0); dep[s] = 1; 
    while (!Q.empty()) {
        int u = Q.front(); Q.pop();
        for (int i = head[u]; ~i; i = e[i].next) {
            int v = e[i].to, w = e[i].w;
            if (w > 0 && !dep[v]) {
                dep[v] = dep[u] + 1;
                Q.push(v); if (v == t) return 1;
            }
        }
    } return 0; 
}

int dfs(int u, int _w) {
    if (!_w || u == t) return _w;
    int s = 0;
    for (int i = head[u]; ~i; i = e[i].next) {
        int v = e[i].to, w = e[i].w;
        if (w > 0 && dep[v] == dep[u] + 1) {
            int di = dfs(v, min(_w - s, w));
            e[i].w -= di; e[i ^ 1].w += di;
            s += di; if (s == _w) break; 
        }
    }
    if (s < _w) dep[u] = 0; 
    return s;
} 

int dinic() { int mf = 0; 
    while (bfs()) mf += dfs(s, INF);
    return mf; 
} 

int n;
int x[maxn], y[maxn], a[maxn], r[maxn];
int g[maxn][maxn], G[maxn][maxn];

int F(int x, int y) { return (x - y) * (x - y); }

int dfn[maxn], low[maxn], bl[maxn], scc, w[maxn], c2;
bool ins[maxn]; stack<int> S; 
void tarjan(int u) { 
    dfn[u] = low[u] = ++c2; ins[u] = 1; S.push(u);
    for (int v = 1; v <= n; ++v) {
        if (!g[u][v] || u == v) continue;
        if (!dfn[v]) tarjan(v), low[u] = min(low[u], low[v]);
        else if (ins[v]) low[u] = min(low[u], dfn[v]);
    }
    if (dfn[u] == low[u]) {
        int t; w[++scc] = 0; 
        do { 
        	t = S.top(); S.pop(); 
           	w[scc] += a[t]; bl[t] = scc;
        } while (t != u);
    }
}

void rebuild() {
    for (int i = 1; i <= scc; ++i)
        for (int j = 1; j <= scc; ++j) G[i][j] = 0;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= n; ++j)
            if (g[i][j]) G[bl[i]][bl[j]] = 1; 
}

void work() { 
    cin >> n;
    fill(head, head + n + 2, -1); c1 = c2 = scc = 0; 
    for (int i = 1; i <= n; ++i) dfn[i] = low[i] = 0; 
    for (int i = 1; i <= n; ++i) cin >> x[i] >> y[i] >> r[i] >> a[i];
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= n; ++j)
            if (F(x[i], x[j]) + F(y[i], y[j]) <= r[i] * r[i]) g[i][j] = 1;
            else g[i][j] = 0; 
    for (int k = 1; k <= n; ++k)
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n; ++j) g[i][j] |= g[i][k] & g[k][j];
    for (int i = 1; i <= n; ++i) if (!dfn[i]) tarjan(i); rebuild();
    s = 0; t = n + 1; int ans = 0; 
    for (int i = 1; i <= scc; ++i) {
        if (w[i] < 0) { Add_edge(i, t, -w[i]); continue; }
        ans += w[i]; Add_edge(s, i, w[i]);
        for (int j = 1; j <= scc; ++j)
            if (w[j] < 0 && G[i][j]) Add_edge(i, j, INF);
    } cout << ans - dinic() << "\n";
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);
    
    int T; cin >> T; 
    while (T--) work();
    return 0;
}