hdu 1853 (费用流 拆点）

// 给定一个有向图，必须用若干个环来覆盖整个图，要求这些覆盖的环的权值最小。 

思路：原图每个点 u 拆为 u 和 u' ，从源点引容量为 1 费用为 0 的边到 u ，从 u' 引相同性质的边到汇点，若原图中存在 (u, v) ，则从 u 引容量为 1 费用为 c(u, v) 的边到 v' 。

其实这里的源模拟的是出度，汇模拟的是入度，因为环中每个点的出度等于入度等于 1 ，那么如果最大流不等于顶点数 n ，则无解；否则，答案就是最小费用。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;

const int maxn = 2e2 + 5;
const int maxm = 2e4 + 5;
const int inf = 0x3f3f3f3f;

struct MCMF
{
    struct Edge
    {
        int v, c, w, next;
    }p[maxm << 1];
    int e, head[maxn], dis[maxn], pre[maxn], cnt[maxn], sumFlow, n;
    bool vis[maxn];
    void init(int nt)
    {
        e = 0, n = nt;
        memset(head, -1, sizeof(head[0]) * (n + 2) );
    }
    void addEdge(int u, int v, int c, int w)
    {
        p[e].v = v, p[e].c = c; p[e].w = w; p[e].next = head[u]; head[u] = e++;
        swap(u, v);
        p[e].v = v, p[e].c = 0; p[e].w = -w; p[e].next = head[u]; head[u] = e++;
    }
    bool spfa(int S, int T)
    {
        queue <int> q;
        for (int i = 0; i <= n; ++i)
            vis[i] = cnt[i] = 0, pre[i] = -1, dis[i] = inf;
        vis[S] = 1, dis[S] = 0;
        q.push(S);
        while (!q.empty())
        {
            int u = q.front(); q.pop();
            vis[u] = 0;
            for (int i = head[u]; i + 1; i = p[i].next)
            {
                int v = p[i].v;
                if (p[i].c && dis[v] > dis[u] + p[i].w)
                {
                    dis[v] = dis[u] + p[i].w;
                    pre[v] = i;
                    if (!vis[v])
                    {
                        q.push(v);
                        vis[v] = 1;
                        if (++cnt[v] > n) return 0;
                    }
                }
            }
        }
        return dis[T] != inf;
    }
    int mcmf(int S, int T)
    {
        sumFlow = 0;
        int minFlow = 0, minCost = 0;
        while (spfa(S, T))
        {
            minFlow = inf + 1;
            for (int i = pre[T]; i + 1; i = pre[ p[i ^ 1].v ])
                minFlow = min(minFlow, p[i].c);
            sumFlow += minFlow;
            for (int i = pre[T]; i + 1; i = pre[ p[i ^ 1].v ])
            {
                p[i].c -= minFlow;
                p[i ^ 1].c += minFlow;
            }
            minCost += dis[T] * minFlow;
        }
        return minCost;
    }
    void build(int nt, int mt)
    {
        init((nt << 1) + 1);
        int u, v, c;
        for (int i = 1; i <= nt; ++i)
        {
            addEdge(0, i, 1, 0);
            addEdge(i + nt, n, 1, 0);
        }
        while (mt--)
        {
            scanf("%d%d%d", &u, &v, &c);
            addEdge(u, v + nt, 1, c);
        }
    }
    void solve(int nt)
    {
        int ans = mcmf(0, n);
        if ( sumFlow != nt)
            printf("-1
");
        else
            printf("%d
", ans);
    }
}my;

int main()
{
    int n, m;
    while (~scanf("%d%d", &n, &m))
    {
        my.build(n, m);
        my.solve(n);
    }
    return 0;
}