最小树形图,测模版....
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.
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.
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.
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
17 6
Problem setter: Hill
/* ***********************************************
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;
}
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TJU 2248. Channel Design 最小树形图
原文：http://blog.csdn.net/ck_boss/article/details/46759345