Ice_cream’s world II

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

無根最小樹形圖

新增一個超級原點，從原點向1-n所有點建立一個權值為sum的有向邊，sum = 所有邊權和加1，然後跑以超級原點為root跑最小樹形圖，最終的答案 = ans-sum,  root 是超級原點指向的點，用邊和點的對應關係可以進行對映確定。

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

using namespace std;
const int maxn = 1e3 + 10;
const int maxm = 4e4 + 10;
const long long INF =  ((1LL) << 60);
const double eps = 1e-8;
#define type long long
int rt;
type in[maxn];
int id[maxn],used[maxn],pre[maxn];

struct edge
{
  int u,v;
  type c;
  edge(int x = 0, int y = 0, type z = 0) : u(x),v(y),c(z) {}
}es[maxm];

type MTG(int root, int n, int m)
{
  type res = 0;
  while(1)
  {
    for(int i = 1; i <= n; i++)  in[i] = INF,id[i] = used[i] = -1;
    for(int i = 0; i < m; i++)
     {
       int u = es[i].u;
       int v = es[i].v;
       if(es[i].c < in[v] && u != v)
       {
         in[v] = es[i].c,pre[v] = u;
         if(u == root) rt = i;
       }
     }//找最小的入邊
    for(int i = 1; i <= n; i++)
     {
       if(i == root) continue;
       res += in[i];
       if(in[i] == INF) return -1;
     }//加入所有的入邊權到答案中
    int cnt = 0;
    for(int i = 1; i <= n; i++)//列舉每個點為起點進行找環
    {
      int v = i;
      while(v != root && used[v] != i && id[v] == -1)  used[v] = i,v = pre[v];
      if(v != root && id[v] == -1)//找到環的時候進行縮點編號
       {
         ++cnt;
         id[v] = cnt;
         for(int u = pre[v]; u != v; u = pre[u]) id[u] = cnt;
       }
    }
    if(cnt == 0) break;//無環則退出
    for(int i = 1; i <= n; i++)
       if(id[i] == -1)  id[i] = ++cnt;

    for(int i = 0; i < m; i++)//更新邊
     {
       int u = es[i].u;
       int v = es[i].v;
       es[i].u = id[u];
       es[i].v = id[v];
       if(u != v)  es[i].c -= in[v]; //更新邊權
     }
     n = cnt;//更新節點數目和根節點編號
     root = id[root];
     }
     return res;
}
int n,m;

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
      type sum = 1;
      for(int i = 0; i < m; i++)
    {
      int u,v;
      type c;
      scanf("%d%d%lld",&u,&v,&c);
      u++,v++;
      sum += c;
      es[i] = edge(u,v,c);
    }

    for(int i = 1; i <= n; i++)
     {
       es[m+i-1] = edge(n+1,i,sum);
     }
    type ans = MTG(n+1,n+1,m+n);
    //cout << ans << endl;
    //printf("Case #%d: ",++cs);
    if(ans >= 2*sum) printf("impossible\n\n");
    else printf("%lld %d\n\n",ans-sum,rt-m);
  }
  return 0;
}