学步园

题目分析：最小树形图模板题。。。。。

码完以后调都没调试，交上去一遍AC。。。。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

#define REP( i , a , b ) for ( int i = a ; i < b ; ++ i )
#define REV( i , a , b ) for ( int i = a - 1 ; i >= b ; -- i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define FOV( i , a , b ) for ( int i = a ; i >= b ; -- i )
#define CLR( a , x ) memset ( a , x , sizeof a )

const int MAXN = 1005 ;
const int MAXE = 40005 ;
const int INF = 0x3f3f3f3f ;

struct Edge {
	int u , v , c ;
} ;

struct Directed_MST {
	Edge E[MAXE] ;
	int in[MAXN] ;
	int idx[MAXN] ;
	int vis[MAXN] ;
	int p[MAXN] ;
	int root ;
	int res ;
	int NV , NE ;
	
	int zhuliu () {
		res = 0 ;
		root = 0 ;
		while ( 1 ) {
			REP ( i , 0 , NV )
				in[i] = INF ;
			REP ( i , 0 , NE ) {
				int u = E[i].u ;
				int v = E[i].v ;
				if ( u != v && in[v] > E[i].c ) {
					in[v] = E[i].c ;
					p[v] = u ;
				}
			}
			REP ( i , 0 , NV )
				if ( in[i] == INF && i != root )
					return 0 ;
			int cnt = 0 ;
			CLR ( vis , -1 ) ;
			CLR ( idx , -1 ) ;
			in[root] = 0 ;
			REP ( i , 0 , NV ) {
				res += in[i] ;
				int v = i ;
				while ( vis[v] != i && idx[v] == -1 && v != root ) {
					vis[v] = i ;
					v = p[v] ;
				}
				if ( idx[v] == -1 && v != root ) {
					for ( int u = p[v] ; u != v ; u = p[u] )
						idx[u] = cnt ;
					idx[v] = cnt ++ ;
				}
			}
			if ( !cnt )
				break ;
			REP ( i , 0 , NV )
				if ( idx[i] == -1 )
					idx[i] = cnt ++ ;
			REP ( i , 0 , NE ) {
				int u = E[i].u ;
				int v = E[i].v ;
				E[i].u = idx[u] ;
				E[i].v = idx[v] ;
				if ( idx[u] != idx[v] )
					E[i].c -= in[v] ;
			}
			NV = cnt ;
			root = idx[root] ;
		}
		return 1 ;
	}
	
	void solve () {
		scanf ( "%d%d" , &NV , &NE ) ;
		REP ( i , 0 , NE )
			scanf ( "%d%d%d" , &E[i].u , &E[i].v , &E[i].c ) ;
		if ( zhuliu () )
			printf ( "%d\n" , res ) ;
		else
			printf ( "Possums!\n" ) ;
	}
} z ;

int main () {
	int T , cas = 0 ;
	scanf ( "%d" , &T ) ;
	while ( T -- ) {
		printf ( "Case #%d: " , ++ cas ) ;
		z.solve () ;
	}
	return 0 ;
}