UVALive - 2197 Paint the Roads(费用流)

题目大意：有n个点，m条边，你的任务是选择其中的一些边，使得每条被选择的边组成一些没有公共边的回路，且每个城市恰好在其中的k个回路上，被选择的边的总权值要求最小

解题思路：k个回路，每个城市都有，表示每个城市的入度和出度都是k，所以以此建边

源点连向每个城市，容量为k，费用0

每个城市连向汇点，容量为k，费用0

边连接两个城市，容量为1，费用为权值

跑最小费用最大流

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
#define N 1010
#define INF 0x3f3f3f3f

struct Edge{
    int from, to, cap, flow ,cost;
    Edge() {}
    Edge(int from, int to, int cap, int flow, int cost):from(from), to(to), cap(cap), flow(flow), cost(cost) {}
};

struct MCMF{
    int n, m, source, sink, flow, cost;
    vector<Edge> edges;
    vector<int> G[N];
    int d[N], f[N], p[N];
    bool vis[N];

    void init(int n) {
        this->n = n;
        for (int i = 0; i <= n; i++)
            G[i].clear();
        edges.clear();
    }

    void AddEdge(int from, int to, int cap, int cost) {
        edges.push_back(Edge(from, to, cap, 0, cost));
        edges.push_back(Edge(to, from, 0, 0, -cost));
        m = edges.size();
        G[from].push_back(m - 2);
        G[to].push_back(m - 1);
    }

    bool BellmanFord(int s, int t, int &flow, int &cost) {
        for (int i = 0; i <= n; i++)
            d[i] = INF;
        memset(vis, 0, sizeof(vis));
        vis[s] = 1; d[s] = 0; f[s] = INF; p[s] = 0;
        queue<int> Q;
        Q.push(s);

        while (!Q.empty()) {
            int u = Q.front();
            Q.pop();
            vis[u] = 0;

            for (int i = 0; i < G[u].size(); i++) {
                Edge &e = edges[G[u][i]];
                if (e.cap > e.flow && d[e.to] > d[u] + e.cost) {
                    d[e.to] = d[u] + e.cost;
                    p[e.to] = G[u][i];
                    f[e.to] = min(f[u], e.cap - e.flow);
                    if (!vis[e.to]) {
                        vis[e.to] = true;
                        Q.push(e.to);
                    }
                }
            }
        }

        if (d[t] == INF)
            return false;

        flow += f[t];
        cost += d[t];

        int u = t;
        while (u != s) {
            edges[p[u]].flow += f[t];
            edges[p[u] ^ 1].flow -= f[t];
            u = edges[p[u]].from;
        }
        return true;
    }

    void Mincost(int s, int t) {
        flow = 0, cost = 0;
        while (BellmanFord(s, t, flow, cost));
    }
};
MCMF mcmf;

int n, m, k;
void solve() {
    scanf("%d%d%d", &n, &m, &k);
    int source = 2 * n, sink = 2 * n + 1;
    mcmf.init(sink);
    for (int i = 0; i < n; i++) {
        mcmf.AddEdge(source, i * 2, k, 0);
        mcmf.AddEdge(i * 2 + 1, sink, k, 0);
    }
    int u, v, c;
    for (int i = 1; i <= m; i++) {
        scanf("%d%d%d", &u, &v, &c);
        mcmf.AddEdge(u * 2, v * 2 + 1, 1, c);
    }
    mcmf.Mincost(source, sink);
    if (mcmf.flow == k * n) printf("%d\n", mcmf.cost);
    else printf("-1\n");

}

int main() {
    int test;
    scanf("%d", &test);
    while (test--) solve();
    return 0;
}

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