HDU 2121 Ice_cream’s world II 最小树形图（不定根）

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Ice_cream’s world II

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2290    Accepted Submission(s): 543

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

 

Recommend

威士忌

 

给你城市个数和可以连接的路径及花费，路径是单向传递的，让你求连接所有城市所需要的最小花费及此时的根。

首先虚拟出一个根，让这个根和所有的节点相连，权值为无穷大或者所有边的权值+1，记为max。然后用祝刘算法进行计算，算出的结果减去max，如果得出的结果为-1或者比max还要大，则不存在最小树形图。求最小根时，在找最小入弧时，如果这条弧的起点是虚拟根，那么这条弧的终点就是要求的根。

#include<stdio.h>
#include<string.h>
#include<math.h>
#define M 10007
#define inf 0x3f3f3f
using namespace std;
int pre[M],vis[M],id[M];
int in[M];
int n,m,ansi;

struct Node//建立邻接表
{
    int u,v;
    int cost;
}E[M*M+5];

int direct_mst(int root,int nv,int ne)
{
    int ret=0;
    while(true)
    {
        //找最小入边
        for(int i=0;i<nv;i++)
            in[i]=inf;
        for(int i=0;i<ne;i++)
        {
            int u=E[i].u;
            int v=E[i].v;
            if(E[i].cost<in[v]&&u!=v)
            {
                pre[v]=u;
                if(u==root)
                    ansi=i;
                in[v]=E[i].cost;
            }
        }
        for(int i=0;i<nv;i++)
        {
            if(i==root)continue;
            if(in[i]==inf)return -1;
        }
        //找环
        int cntnode=0;
        memset(id,-1,sizeof(id));
        memset(vis,-1,sizeof(vis));
        in[root]=0;
        for(int i=0;i<nv;i++)//标记每个环
        {
            ret+=in[i];
            int 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]=cntnode;
                }
                id[v]=cntnode++;
            }
        }
        if(cntnode==0)break;//无环
        for(int i=0;i<nv;i++)
            if(id[i]==-1)
                id[i]=cntnode++;
        //缩点，重新标记
        for(int i=0;i<ne;i++)
        {
            int v=E[i].v;
            E[i].u=id[E[i].u];
            E[i].v=id[E[i].v];
            if(E[i].u!=E[i].v)
            {
                E[i].cost-=in[v];
            }
        }
        nv=cntnode;
        root=id[root];
    }
    return ret;
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        int u,v,w,sum=0;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            E[i].u=++u;
            E[i].v=++v;
            if(u!=v)
                E[i].cost=w;
            else
                E[i].cost=inf;
            sum+=w;
        }
        sum++;
        for(int i=0;i<n;i++)
        {
            E[m+i].u=0;
            E[m+i].v=i+1;
            E[m+i].cost=sum;
        }
        int ans=direct_mst(0,n+1,m+n);//n代表点的个数，m代表边的个数
        if(ans==-1||ans-sum>=sum)
            printf("impossible\n");//无法生成最小树形图
        else
            printf("%d %d\n",ans-sum,ansi-m);
        printf("\n");
    }
    return 0;
}