TJU 2248. Channel Design 最小树形图

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2248.   Channel Design

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Time Limit: 1.0 Seconds   Memory Limit: 65536K

Total Runs: 2199   Accepted Runs: 740

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We need irrigate our farms, but there is only one source of water nearby. So we need build some water channels with minimum cost.

In Figure (a), V₁ indicates the source of water. Other N-1 nodes in the Figure indicate the farms we need to irrigate. An edge represents you
can build a channel between the two nodes, to irrigate the target. The integers indicate the cost of a channel between two nodes.

Figure (b) represents a design of channels with minimum cost.

Input

There are multiple cases, the first line of each case contains two integers N and M (2 ≤ N ≤ 100; 1 ≤ M ≤ 10000), N shows the number of nodes. The following M lines,
each line contains three integers i j c_ij, means we can build a channel from node V_i to node V_j, which cost c_ij.
(1 ≤ i, j ≤ N; i ≠ j; 1 ≤ c_ij ≤ 100)

The source of water is always V₁.

The input is terminated by N = M = 0.

Output

For each case, output a single line contains an integer represents the minimum cost.

If no design can irrigate all the farms, output "impossible" instead.

Sample Input

5 8
1 2 3
1 3 5
2 4 2
3 1 5
3 2 5
3 4 4
3 5 7
5 4 3
3 3
1 2 3
1 3 5
3 2 1
0 0

Sample Output

17
6

Problem setter: Hill

Source: TJU Contest August
2006

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/* ***********************************************
Author        :CKboss
Created Time  :2015年07月04日 星期六 23时35分05秒
File Name     :TJU2248.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

const int INF=0x3f3f3f3f;
const int maxn=110;

int n,m;

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

Edge edge[maxn*maxn];
int pre[maxn],id[maxn],vis[maxn],in[maxn];

int zhuliu(int root,int n,int m,Edge edge[])
{
	int res=0,u,v;
	while(true)
	{
		for(int i=0;i<n;i++) in[i]=INF;
		for(int i=0;i<m;i++)
		{
			if(edge[i].u!=edge[i].v&&edge[i].cost<in[edge[i].v])
			{
				pre[edge[i].v]=edge[i].u;
				in[edge[i].v]=edge[i].cost;
			}
		}
		for(int i=0;i<n;i++)
			if(i!=root&&in[i]==INF) return -1;
		int tn=0;
		memset(id,-1,sizeof(id));
		memset(vis,-1,sizeof(vis));
		in[root]=0;
		for(int i=0;i<n;i++)
		{
			res+=in[i];
			v=i;
			while(vis[v]!=i&&id[v]==-1&&v!=root)
			{
				vis[v]=i; v=pre[v];
			}
			if(v!=root&&id[v]==-1)
			{
				for(int u=pre[v];u!=v;u=pre[u])
					id[u]=tn;
				id[v]=tn++;
			}
		}
		if(tn==0) break;
		for(int i=0;i<n;i++)
			if(id[i]==-1) id[i]=tn++;
		for(int i=0;i<m;)
		{
			v=edge[i].v;
			edge[i].u=id[edge[i].u];
			edge[i].v=id[edge[i].v];
			if(edge[i].u!=edge[i].v)
				edge[i++].cost-=in[v];
			else
				swap(edge[i],edge[--m]);
		}
		n=tn;
		root=id[root];
	}
	return res;
}

int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);

	while(scanf("%d%d",&n,&m)!=EOF)
	{
		if(n==0&&m==0) break;
		for(int i=0;i<m;i++)
		{
			int u,v,w;
			scanf("%d%d%d",&u,&v,&w); u--; v--;
			edge[i].u=u; edge[i].v=v; edge[i].cost=w;
		}
		int ans=zhuliu(0,n,m,edge);
		if(ans==-1) puts("impossible");
		else printf("%d\n",ans);
	}

    return 0;
}