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

Ice_cream’s world II

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5665    Accepted Submission(s): 1437

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 

 

题意：

就是不确定根求最小树形图，输出根。

POINT：

原图中所有权值和和x，建一个虚根连上所有点，长度为sum=x+1.

注意sum不能不加一，我们可以假设存在一种情况有两个点，0条边，那么此时会有两条从超级源点出发的长度为0的边，那么此时无法判断是不是满足条件(0>2*0不满足，会认为是满足条件的，实际上0条边是不满足条件的)

然后ans>=sum*2即无解。然后不用删边，用来记录真根。

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <string>
typedef long long LL;
using namespace std;
const LL maxn = 1111;
const LL maxm = 11111;
const LL inf = 999999999999999;

struct edge
{
    LL u,v,w;
}len[maxm];
LL in[maxn],pre[maxn],vis[maxn],id[maxn];
LL pos;
LL zhuliu(LL root,LL n,LL m)//[0-n]点 虚根为0
{
    LL res=0,u,v;
    while(1)
    {
        LL cn=0;
        for(LL i=0;i<=n;i++) in[i]=inf;
        for(LL i=1;i<=m;i++)
        {
            u=len[i].u,v=len[i].v;
            if(u!=v&&in[v]>len[i].w)
            {
                in[v]=len[i].w;
                pre[v]=u;
                if(u==root)
                    pos=i;
            }
        }
        in[root]=0;
        for(LL i=0;i<=n;i++) if(in[i]==inf) return inf;
        memset(id,-1,sizeof id);
        memset(vis,-1,sizeof vis);
        for(LL 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(id[v]==-1&&v!=root)
            {
                for(u=pre[v];u!=v;u=pre[u])
                {
                    id[u]=cn;
                }
                id[u]=cn;
                cn++;
            }
        }
        if(cn==0) break;
        for(LL i=0;i<=n;i++) if(id[i]==-1) id[i]=cn++;
        for(LL i=1;i<=m;i++)
        {
            v=len[i].v;
            len[i].u=id[len[i].u];
            len[i].v=id[len[i].v];
            if(len[i].u!=len[i].v)
            {
                len[i].w-=in[v];
            }
        }
        n=cn-1;
        root=id[root];
    }
    return res;
}
int main()
{
    LL n,m;
    while(~scanf("%lld %lld",&n,&m))
    {
        LL sum=1;
        for(LL i=1;i<=m;i++)
        {
            LL u,v,w;scanf("%lld %lld %lld",&u,&v,&w);
            u++,v++;
            len[i].u=u,len[i].v=v,len[i].w=w;
            sum+=w;
        }
        for(LL i=1;i<=n;i++)
        {
            len[i+m].u=0,len[i+m].v=i,len[i+m].w=sum;
        }
        LL ans=zhuliu(0,n,n+m);
        if(ans>=2*sum)
        {
            printf("impossible\n");
        }
        else
        {
            printf("%lld %lld\n",ans-sum,pos-m-1);
        }
        printf("\n");
    }
}