【HDU】3861 The King’s Problem 强连通缩点+有向图最小路径覆盖

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 ;
}