TJU 2248. Channel Design 最小树形图

最小树形图,測模版....

2248. Channel Design

Time Limit: 1.0 Seconds Memory Limit: 65536K

Total Runs: 2199 Accepted Runs: 740

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 cij, means we can build a channel from node Vi to node Vj, which cost cij.
(1 ≤ i, j ≤ N; i ≠ j; 1 ≤ cij ≤ 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

Submit List
Runs Forum Statistics

/* ***********************************************
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;
}