网络流24题（03）最小路径覆盖问题（二分图匹配 + 最大流）

思路：

1. 把每个点拆分成 2 个点 Xi, Yi，由 s 向 Xi 引弧，Yi 向 t 引弧，如果 Xi, Yj 存在弧则引弧。所有弧的容量均为 1;

2. 这样就构造出来了二分图的模型，然后求最大流即是这个二分图的最大匹配了。路径数 = 点数 - 最大流;

3. 因为如果存在一个匹配边，则被覆盖的点数就会减 1，所以此时路径数就是如 2 中求得;

 

#include <cstdio>

#include <cstring>

#include <cctype>

#include <algorithm>

#include <queue>

#include <vector>

using namespace std;



const int MAXN = 1010;

const int INFS = 0x3FFFFFFF;



struct edge {

    int from, to, cap, flow;

    edge(int _from, int _to, int _cap, int _flow) 

        : from(_from), to(_to), cap(_cap), flow(_flow) {}

};



class Dinic {

public:

    void initdata(int n, int s, int t) {

        this->n = n, this->s = s, this->t = t;

        edges.clear();

        for (int i = 0; i < n; i++)

            G[i].clear();

    }

    void addedge(int u, int v, int cap) {

        edges.push_back(edge(u, v, cap, 0));

        edges.push_back(edge(v, u, 0, 0));

        G[u].push_back(edges.size() - 2);

        G[v].push_back(edges.size() - 1);

    }

    bool BFS() {

        for (int i = 0; i < n; i++)

            vis[i] = false, d[i] = 0;

        queue<int> Q;

        Q.push(s);

        vis[s] = true;

        while (!Q.empty()) {

            int x = Q.front(); Q.pop();

            for (int i = 0; i < G[x].size(); i++) {

                edge& e = edges[G[x][i]];

                if (e.cap > e.flow && !vis[e.to]) {

                    vis[e.to] = true;

                    d[e.to] = d[x] + 1;

                    Q.push(e.to);

                }

            }

        }

        return vis[t];

    }

    int DFS(int x, int aug) {

        if (x == t || aug == 0) return aug;

        int flow = 0;

        for (int i = 0; i < G[x].size(); i++) {

            edge& e = edges[G[x][i]];

            if (d[e.to] == d[x] + 1) {

                int f = DFS(e.to, min(aug, e.cap - e.flow));

                if (f <= 0) continue;

                e.flow += f;

                edges[G[x][i]^1].flow -= f;

                flow += f;

                aug -= f;

                if (aug == 0) break;

            } 

        }

        return flow;

    }

    int maxflow() {

        int flow = 0;

        while (BFS()) {

            flow += DFS(s, INFS);

        }

        return flow;

    }

    void print(int x) {

        vis[x] = true;

        int num = (n-2)/2;

        for (int i = 0; i < G[x].size(); i++) {

            edge& e = edges[G[x][i]];

            if (!vis[e.to] && e.to != t && e.flow == 1) {

                printf(" %d", e.to - num); print(e.to - num);

            }

        }

    }

    void printpath() {

        int flow = maxflow();

        memset(vis, false, sizeof(vis));

        int num = (n-2)/2;

        for (int i = 1; i <= num; i++) {

            if (!vis[i]) {

                printf("%d", i);

                print(i); printf("\n");

            }

        }

        printf("%d\n", num - flow);

    }

private:

    vector<edge> edges;

    vector<int> G[MAXN];

    int n, s, t, d[MAXN];

    bool vis[MAXN];

};



Dinic dinic;



int main() {

    int n, m;

    scanf("%d%d", &n, &m);

    int s = 0, t = 2*n + 1;

    dinic.initdata(t+1, s, t);

    for (int i = 1; i <= n; i++) {

        dinic.addedge(s, i, 1);

        dinic.addedge(i + n, t, 1);

    }

    while (m--) {

        int u, v;

        scanf("%d%d", &u, &v);

        dinic.addedge(u, v + n, 1);

    }

    dinic.printpath();

    return 0;

}