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  • 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 ≤ ij ≤ Ni ≠ 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
    ",ans);
    	}
        
        return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/mfrbuaa/p/5388191.html
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