UVALive - 2197 Paint the Roads(费用流)

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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 0x3f3f3f3fstruct 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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