Everyone is strong enough to share their support.

Topic:

_Given a directed graph G=(V,E).Set P to be a set of G's simple paths with disjoint vertices.If each vertex in V happens to be on a path of P, P is said to be a path coverage of G.A path in P can start from any vertex of V and have an arbitrary length, especially 0.The minimum path coverage of G is the path coverage of G with the least number of paths.

Questions:

_is transformed into a problem of large drop matching in a bipartite graph, where each point u is split into a point u x on the left and a point uy on the right, and then (u x, vy) is connected to an arc with a flow rate of x(x <[1,+ <)] for an edge (u,v).

_Finds its drop size match and subtracts the total number of points to get the minimum path coverage.

Prove:

If you want to output the solution, just look for the edge with 0 traffic in it...See code for details

# include <bits/stdc++.h>

# define N 2010

class Network  {
private :
    struct edge  {
        int to, w, nxt ;
        edge ( ) {        }
        edge ( int to, int w, int nxt ) : to ( to ), w ( w ), nxt ( nxt ) {        }  
    } g [60010 << 1] ;

    int head [N], cur [N], ecnt ;
    int S, T , dep [N] ;

    inline int dfs ( int u, int a )  {
        if ( u == T || ! a )  return a ;
        int flow = 0, v, f ;
        for ( int& i = cur [u] ; i ; i = g [i].nxt )  {
            v = g [i].to ;
            if ( dep [v] == dep [u] + 1 )  {
                f = dfs ( v, std :: min ( g [i].w, a - flow ) ) ;
                g [i].w -= f, g [i ^ 1].w += f ;
                flow += f ;
                if ( a == flow )  return a ;
            }
        }
        if ( ! flow )  dep [u] = -1 ;
        return flow ;
    }

    inline bool bfs ( int S, int T )  {
        static std :: queue < int > q ;
        memset ( dep, 0, sizeof ( int ) * ( T + 1 ) ) ;
        dep [S] = 1 ;
        q.push ( S ) ;
        while ( ! q.empty ( ) )  {
            int u = q.front ( ) ;  q.pop ( ) ;
            for ( int i = head [u] ; i ; i = g [i].nxt )  {
                int& v = g [i].to ;
                if ( g [i].w &&  ! dep [v] )  {
                    dep [v] = dep [u] + 1 ;
                    q.push ( v ) ;
                }
            }
        }
        return dep [T] ;
    }
public :
    Network ( )  {    ecnt = 1 ; }

    inline void add_edge ( int u, int v, int w )  {
        g [++ ecnt] = edge ( v, w, head [u] ) ;     head [u] = ecnt ;
        g [++ ecnt] = edge ( u, 0, head [v] ) ;     head [v] = ecnt ;
    }

    inline int dinic ( int S, int T )  {
        this -> S = S, this -> T = T ;
        int rt = 0 ;
        while ( bfs ( S, T ) )    {
            memcpy ( cur, head, sizeof ( int ) * ( T + 1 ) ) ; 
            rt += dfs ( S, 0x3f3f3f3f ) ;
        }
        return rt ;
    }

    void display ( int n, int S, int T )  {

        int ans = n - dinic ( S, T ) ;

        static std :: bitset < N > vis ;
        vis.reset ( ) ;
        for ( int i = 1 ; i <= n ; ++ i )
            if ( ! vis.test ( i ) )  {
                int curr = i ;
                for ( bool found = 0 ; ; found = 0 )  {
                    printf ( "%d ", curr ) ;
                    vis.set ( curr ) ;
                    for ( int i = head [curr] ; i ; i = g [i].nxt )  {
                        int& v = g [i].to ;
                        if ( g [i].w == 0 && v > n && v <= 2 * n )  {
                            curr = v - n ;
                            found = 1 ;
                            break ;
                        }
                    }
                    if ( found == 0 )  {
                        puts ( "" ) ;
                        break ;
                    }
                }
            }

        printf ( "%d\n", ans ) ;
    }
} Lazer ;

int main ( )  {
    int n, m ;
    scanf ( "%d%d", & n, & m ) ;

    const int S = n * 2 + 1, T = n * 2 + 2 ;

    for ( int i = 1 ; i <= n ; ++ i )  {
        Lazer.add_edge ( S, i, 1 ) ;
        Lazer.add_edge ( i + n, T, 1 ) ;
    }

    while ( m -- )  {
        static int u, v ;
        scanf ( "%d%d", & u, & v ) ;
        Lazer.add_edge ( u, v + n, 1 ) ;
    }

    Lazer.display ( n, S, T ) ;
    return 0 ;
}