本文主要是介绍【HDU】3861 The King’s Problem 强连通缩点+有向图最小路径覆盖，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
传送门：【HDU】3861 The King’s Problem
题目分析：首先强连通缩点，因为形成一个环的王国肯定在一条路径中，这样才能保证拆的少。
然后缩点后就是DAG图了，由于题目要求的是最小路径覆盖，那么二分匹配即可。
代码如下：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <time.h>
#include <stdlib.h>
using namespace std ;#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REPV( i , a , b ) for ( int i = a ; i >= b ; -- i )
#define CLR( a , x ) memset ( a , x , sizeof a )const int MAXN = 5005 ;
const int MAXE = 100005 ;
const int MAXQ = 1000005 ;
const int INF = 0x3f3f3f3f ;struct Edge {int v , n ;Edge () {}Edge ( int v , int n ) : v ( v ) , n ( n ) {}
} ;struct CC {Edge E[MAXE] ;int H[MAXN] , cntE ;int low[MAXN] , dfn[MAXN] , dfs_clock ;int S[MAXN] , top ;int scc[MAXN] , scc_cnt ;int n , m ;void init () {cntE = dfs_clock = scc_cnt = top = 0 ;CLR ( H , -1 ) ;CLR ( scc , 0 ) ;CLR ( dfn , 0 ) ;}void addedge ( int u , int v ) {E[cntE] = Edge ( v , H[u] ) ;H[u] = cntE ++ ;}void Tarjan ( int u ) {low[u] = dfn[u] = ++ dfs_clock ;S[top ++] = u ;for ( int i = H[u] ; ~i ; i = E[i].n ) {int v = E[i].v ;if ( !dfn[v] ) {Tarjan ( v ) ;low[u] = min ( low[u] , low[v] ) ;}else if ( !scc[v] )low[u] = min ( low[u] , dfn[v] ) ;}if ( low[u] == dfn[u] ) {++ scc_cnt ;while ( 1 ) {int v = S[-- top] ;scc[v] = scc_cnt ;if ( v == u )break ;}}}void input () {int u , v ;scanf ( "%d%d" , &n , &m ) ;REP ( i , m ) {scanf ( "%d%d" , &u , &v ) ;addedge ( u , v ) ;}}void find_scc () {REPF ( i , 1 , n )if ( !dfn[i] )Tarjan ( i ) ;}void solve () {init () ;input () ;find_scc () ;}
} c ;struct Match {Edge E[MAXE] ;int H[MAXN] , cntE ;bool vis[MAXN] ;int Lx[MAXN] , Ly[MAXN] ;int dx[MAXN] , dy[MAXN] ;int Q[MAXQ] , head , tail ;int dis ;int n ;void init () {cntE = 0 ;CLR ( H , -1 ) ;}void addedge ( int u , int v ) {E[cntE] = Edge ( v , H[u] ) ;H[u] = cntE ++ ;}int Hopcroft_Karp () {CLR ( dx , -1 ) ;CLR ( dy , -1 ) ;head = tail = 0 ;dis = INF ;REPF ( i , 1 , n )if ( Lx[i] == -1 ) {Q[tail ++] = i ;dx[i] = 0 ;}while ( head != tail ) {int u = Q[head ++] ;if ( dx[u] >= dis )continue ;for ( int i = H[u] ; ~i ; i = E[i].n ) {int v = E[i].v ;if ( dy[v] == -1 ) {dy[v] = dx[u] + 1 ;if ( Ly[v] == -1 )dis = dy[v] ;else {dx[Ly[v]] = dy[v] + 1 ;Q[tail ++] = Ly[v] ;}}}}return dis != INF ;}int find ( int u ) {for ( int i = H[u] ; ~i ; i = E[i].n ) {int v = E[i].v ;if ( dy[v] == dx[u] + 1 && !vis[v] ) {vis[v] = 1 ;if ( ~Ly[v] && dy[v] == dis )continue ;if ( Ly[v] == -1 || find ( Ly[v] ) ) {Lx[u] = v ;Ly[v] = u ;return 1 ;}}}return 0 ;}int match () {int ans = 0 ;CLR ( Lx , -1 ) ;CLR ( Ly , -1 ) ;while ( Hopcroft_Karp () ) {CLR ( vis , 0 ) ;REPF ( i , 1 , n )if ( Lx[i] == -1 )ans += find ( i ) ;}return ans ;}void solve () {init () ;n = c.scc_cnt ;REPF ( u , 1 , c.n )for ( int i = c.H[u] ; ~i ; i = c.E[i].n ) {int v = c.E[i].v ;if ( c.scc[u] != c.scc[v] )addedge ( c.scc[u] , c.scc[v] ) ;}int ans = match () ;printf ( "%d\n" , n - ans ) ;}
} e ;int main () {int T ;scanf ( "%d" , &T ) ;while ( T -- ) {c.solve () ;e.solve () ;}return 0 ;
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