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

Ice_cream’s world II

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3115    Accepted Submission(s): 737

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

 

设一个虚根 ， 给它到所有点弄一条边，权值足够大 。

不进行删边操作，记录边作为ans2，然后就是 ans2 - m 就是最后要输出的答案了 。

 

 

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <vector>

using namespace std ;
typedef long long LL ;
const double inf = 1e9+7;
const int N = 1010;
const int M = 40010;

struct edge {
    int u , v ;
    LL w;
}e[M];
int n , m , ans2 ;
int pre[N] , id[N] , vis[N] ;
LL in[N] ;
LL zhuliu( int root , int n , int m ) {
    LL res = 0 ; int u , v ;
    while(1) {
        for( int i = 0 ; i < n ; ++i ) in[i] = inf ;

        for( int i = 0 ; i < m ; ++i )
            if( e[i].u != e[i].v && e[i].w < in[e[i].v] ) {
                pre[e[i].v] = e[i].u;
                if( e[i].u == root ) ans2 = i ;
                in[e[i].v] = e[i].w;
            }
        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 ; // no circle
        for( int i = 0 ; i < n ; ++i ) if( id[i] == -1 ) {
            id[i] = tn++ ;
        }
        for( int i = 0 ; i < m ; ++i ){
            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].w -= in[v];
        }
        n = tn ;
        root = id[root];
    }
    return res ;
}
int main ()  {
//    freopen("in.txt","r",stdin);
    int _ , cas = 1 ;
    while( ~scanf("%d%d",&n,&m) ) {
        LL sum = 1 ;
        for( int i = 0 ; i < m ; ++i ) {
            scanf("%d%d%I64d",&e[i].u,&e[i].v,&e[i].w);
            sum += e[i].w;
        }
        int L = m ;
        for( int i = 0 ; i < n ; ++i ) {
            e[L].u = n , e[L].v = i , e[L++].w = sum ;
        }
        LL ans = zhuliu( n , n + 1 , L );
        if( ans == -1 || ans >= 2 * sum )
            puts("impossible");
        else {
            ans -= sum ;
            printf("%I64d %d
",ans,ans2 - m);
        }
        puts("");
    }
}