飞行员配对方案问题

传送门

读完题,就知道这就是个裸的二分图皮配,然后,我还是不说匈牙利,(因为我真的不会啊!)

所以,我还是用了喜闻乐见的 $Dinic$ 并且跑的也不慢.唯一难点就是输出方案了吧…输出方案用最后的连通性判断,这题就没了….

$Code:$

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

const int N = 1e5 + 5 ;
const int INF = 1061109567 ;

using std::queue ;

struct edge { int to , next , flow ; } e[(N<<1)] ;

int tot = 1 , head[N] , flor[N] , s , t , n , m , x , y ;

inline void build (int u , int v ,int w) {
    e[++tot].next = head[u] ; e[tot].to = v ;
    e[tot].flow = w ; head[u] = tot ; return;
}

queue < int > q ;
inline bool bfs ( int cur ) {
    while ( ! q.empty () ) q.pop () ;
    memset ( flor , false , sizeof ( flor ) ) ;
    q.push ( cur ) ; flor[cur] = 1 ;
    while ( ! q.empty () ) {
        int j = q.front () ; q.pop () ;
        for (int i = head[j] ; i ; i = e[i].next) {
            int k = e[i].to ;
            if ( ! flor[k] && e[i].flow ) {
                flor[k] = flor[j] + 1 ;
                if ( k == t ) return flor[t] ;
                q.push ( k ) ;
            }
        }
    }
    return flor[t] ;
}

inline int dfs (int cur , int dist) {
    if ( cur == t ) return dist ;
    for (int i = head[cur] ; i ; i = e[i].next) {
        int k = e[i].to ;
        if ( flor[k] == flor[cur] + 1 && e[i].flow ) {
            int now = dfs ( k , std::min ( e[i].flow , dist ) ) ;
            if ( now != 0 ) {
                e[i].flow -= now ;
                e[i^1].flow += now ;
                return now ;
            }
        }
    }
    return 0 ;
}

inline int Dinic () {
    int ans = 0 ;
    while ( bfs ( s ) )
        while ( int now = dfs ( s , INF ) )
            ans += now ;
    return ans ;
}

int main () {
    scanf ("%d%d" , & m , & n ) ; s = 0 ; t = n + 1 ;
    for (int i = 1 ; i <= m ; ++ i) build ( s , i , 1 ) , build ( i , s , 0 ) ; // 连外籍
    for (int i = m + 1 ; i <= n ; ++ i) build ( i , t , 1 ) , build ( t , i , 0 ) ; // 连皇家
    while ( scanf ("%d%d" , & x , & y ) && x != - 1 && y != - 1 )
        build ( x , y , 1 ) , build ( y , x , 0 ) ;
    int ans = Dinic () ;
    if ( ! ans ) { puts ("No Solution!") ; return 0 ; }
    printf ("%d\n" , ans  ) ;
    for (int i = 2 ; i <= tot ; i += 2)
        if ( e[i].to != s && e[i^1].to != s )
            if ( e[i].to != t && e[i^1].to != t )
                if ( e[i^1].flow != 0 ) printf ("%d %d\n" , e[i^1].to , e[i].to ) ;
    system ("pause") ; return 0 ;
}