朱刘算法 最小树形图无根树:

建立一个虚拟的根节点,向所有节点连边,权值为其他所有边的权值和+1

在求最小树形图的时候,记录和虚拟的根相连的是哪个节点

在这题中,边是从小往大加的所以直接记录的是相连的是第几号边....

Ice_cream’s world II

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

After awarded lands to ACMers, the queen want to choose a city be her capital. This is an important event in ice_cream world, and it also a very difficult problem, because the world have N cities and M roads, every road was directed. Wiskey is a chief engineer
in ice_cream world. The queen asked Wiskey must find a suitable location to establish the capital, beautify the roads which let capital can visit each city and the project’s cost as less as better. If Wiskey can’t fulfill the queen’s require, he will be punishing.

Input

Every case have two integers N and M (N<=1000, M<=10000), the cities numbered 0…N-1, following M lines, each line contain three integers S, T and C, meaning from S to T have a road will cost C.

Output

If no location satisfy the queen’s require, you must be output “impossible”, otherwise, print the minimum cost in this project and suitable city’s number. May be exist many suitable cities, choose the minimum number city. After every case print one blank.

Sample Input

3 1
0 1 1

4 4
0 1 10
0 2 10
1 3 20
2 3 30

Sample Output

impossible

40 0

Author

Wiskey

Source

HDU 2007-10 Programming Contest_WarmUp

/* ***********************************************
Author        :CKboss
Created Time  :2015年07月07日 星期二 16时55分48秒
File Name     :HDOJ2121.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=1100;

int n,m;

struct Edge
{
	int u,v,cost;
}edge[maxn*maxn];

int pre[maxn],id[maxn],vis[maxn];
int in[maxn];
int Root;

int zhuliu(int root,int n,int m,Edge edge[])
{
	int res=0;
	int 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;
				if(edge[i].u==root) Root=i;
			}
		}
		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;i++)
		{
			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];
		}

		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)
	{
		int sum=0;
		int en=0;
		Root=INF;

		for(int i=0;i<m;i++)
		{
			scanf("%d%d%d",&edge[en].u,&edge[en].v,&edge[en].cost);
			sum+=edge[en++].cost;
		}

		sum++;

		/// link with virtual root
		for(int i=0;i<n;i++)
		{
			edge[en].u=n; edge[en].v=i; edge[en].cost=sum;
			en++;
		}

		int lens = zhuliu(n,n+1,en,edge);

		if(lens>=2*sum||lens==-1) puts("impossible");
		else printf("%d %d\n",lens-sum,Root-m);

		putchar(10);
	}

    return 0;
}

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