【HDU】3072 Intelligence System 强连通缩点+贪心

Intelligence System

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1301    Accepted Submission(s): 579

Problem Description

After a day, ALPCs finally complete their ultimate intelligence system, the purpose of it is of course for ACM ... ...
Now, kzc_tc, the head of the Intelligence Department (his code is once 48, but now 0), is sudden obtaining important information from one Intelligence personnel. That relates to the strategic direction and future development of the situation of ALPC. So it need for emergency notification to all Intelligence personnel, he decides to use the intelligence system (kzc_tc inform one, and the one inform other one or more, and so on. Finally the information is known to all).
We know this is a dangerous work. Each transmission of the information can only be made through a fixed approach, from a fixed person to another fixed, and cannot be exchanged, but between two persons may have more than one way for transferring. Each act of the transmission cost Ci (1 <= Ci <= 100000), the total cost of the transmission if inform some ones in our ALPC intelligence agency is their costs sum.
Something good, if two people can inform each other, directly or indirectly through someone else, then they belong to the same branch (kzc_tc is in one branch, too!). This case, it’s very easy to inform each other, so that the cost between persons in the same branch will be ignored. The number of branch in intelligence agency is no more than one hundred.
As a result of the current tensions of ALPC’s funds, kzc_tc now has all relationships in his Intelligence system, and he want to write a program to achieve the minimum cost to ensure that everyone knows this intelligence.
It's really annoying!

 

Input

There are several test cases.
In each case, the first line is an Integer N (0< N <= 50000), the number of the intelligence personnel including kzc_tc. Their code is numbered from 0 to N-1. And then M (0<= M <= 100000), the number of the transmission approach.
The next M lines, each line contains three integers, X, Y and C means person X transfer information to person Y cost C.

 

Output

The minimum total cost for inform everyone.
Believe kzc_tc’s working! There always is a way for him to communicate with all other intelligence personnel.

 

Sample Input

   
   
   
   

    
    
    
    3 3 0 1 100 1 2 50 0 2 100 3 3 0 1 100 1 2 50 2 1 100 2 2 0 1 50 0 1 100
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    150 100 50
   
   
   
   

 

Source

2009 Multi-University Training Contest 17 - Host by NUDT

传送门：【HDU】3072 Intelligence System

题目大意：
给你一副有向图，其中每条边都有一个权值，现在你的目的是选择一些边使得图连通并且权和最小，其中，如果选择后使得一些点强连通，则在强连通分量上的边权值为0。

题目分析：首先强连通缩点，因为一个强连通内的边不计入权和内，所以就不用管了，然后根据题意，我们可以为每个点贪心的选择一条权值最小的边作为入边（因为缩点以后是一个DAG图，所以没有入度的点就不用选择了），最后所有选择的边的权和就是答案。

PS：因为将continue 打成 return。。导致我错的飞起来QUQ
代码如下：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std ;

#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )

const int MAXN = 50005 ;
const int MAXE = 100005 ;
const int MAXQ = 100005 ;
const int MAXM = 105 ;
const int INF = 0x3f3f3f3f ;

struct Edge {
	int v , n , w ;
	Edge ( int v = 0 , int n = 0 , int w = 0 ) : v ( v ) , n ( n ) , w ( w ) {}
} ;

struct SCC {
	Edge edge[MAXE] ;
	int adj[MAXN] , cntE ;
	int Dfn[MAXN] , Low[MAXN] ;
	int dfs_clock , scc_cnt ;
	int S[MAXN] , top ;
	bool ins[MAXN] ;
	int scc[MAXN] ;
	
	void init () {
		top = 0 ;
		cntE = 0 ;
		scc_cnt = 0 ;
		dfs_clock = 0 ;
		CLEAR ( Dfn , 0 ) ;
		CLEAR ( ins , 0 ) ;
		CLEAR ( adj , -1 ) ;
	}
	
	void addedge ( int u , int v , int w ) {
		edge[cntE] = Edge ( v , adj[u] , w ) ;
		adj[u] = cntE ++ ;
	}
	
	void Tarjan ( int u ) {
		Dfn[u] = Low[u] = ++ dfs_clock ;
		ins[u] = true ;
		S[top ++] = u ;
		for ( int i = adj[u] ; ~i ; i = edge[i].n ) {
			int v = edge[i].v ;
			if ( !Dfn[v] ) {
				Tarjan ( v ) ;
				Low[u] = min ( Low[u] , Low[v] ) ;
			}
			else if ( ins[v] )
				Low[u] = min ( Low[u] , Dfn[v] ) ;
		}
		if ( Dfn[u] == Low[u] ) {
			++ scc_cnt ;
			while ( 1 ) {
				int v = S[-- top] ;
				scc[v] = scc_cnt ;
				ins[v] = 0 ;
				if ( v == u )
					break ;
			}
		}
	}
	
	void find_SCC ( int n ) {
		REP ( i , n )
			if( !Dfn[i] )
				Tarjan ( i ) ;
	}
} ;

SCC C ;
int cost[MAXM] ;

void work () {
	int n , m ;
	int u , v , w ;
	while ( ~scanf ( "%d%d" , &n , &m ) ) {
		C.init () ;
		while ( m -- ) {
			scanf ( "%d%d%d" , &u , &v , &w ) ;
			C.addedge ( u , v , w ) ;
		}
		C.find_SCC ( n ) ;
		if ( C.scc_cnt == 1 ) {
			printf ( "0\n" ) ;
			continue ;
		}
		CLEAR ( cost , INF ) ;
		REP ( u , n )
			for ( int i = C.adj[u] ; ~i ; i = C.edge[i].n ) {
				int v = C.edge[i].v ;
				if ( C.scc[u] != C.scc[v] )
					cost[C.scc[v]] = min ( cost[C.scc[v]] , C.edge[i].w ) ;
			}
		int ans = 0 ;
		REPF ( i , 1 , C.scc_cnt )
			if ( cost[i] != INF )
				ans += cost[i] ;
		printf ( "%d\n" , ans ) ;
	}
}
	

int main () {
	work () ;
	return 0 ;
}