TJU 2248. Channel Design 最小树形图

最小树形图,测模版....

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 81 2 31 3 52 4 23 1 53 2 53 4 43 5 75 4 33 31 2 31 3 53 2 10 0

Sample Output

176

Problem setter: Hill

Source: TJU Contest August 2006

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/* ***********************************************Author        :CKbossCreated 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];elseswap(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;}