HDU 2121 Ice_cream’s world II

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

#include<cmath>
#include<vector>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define rep(i,j,k) for(int i=j;i<=k;i++)
#define loop(i,j,k) for (int i=j;i!=-1;i=k[i])
#define inone(x) scanf("%d",&x)
#define intwo(x,y) scanf("%d%d",&x,&y)
#define inthr(x,y,z) scanf("%d%d%d",&x,&y,&z)
const int N = 1e5+ 10;
const int INF = 0x7FFFFFFF;
int n, m, ans, who;
int x[N], y[N], z[N];
int pre[N], in[N], v[N], f[N];

int zhuliu(int rt)
{
  long long res = who = 0;
  while (true)
  {
    rep(i, 0, n) in[i] = INF;
    rep(i, 1, m)
    {
      if (z[i] < in[y[i]] && x[i] != y[i])
      {
        pre[y[i]] = x[i], in[y[i]] = z[i];
        if (x[i] == rt) who = i;
      }
    }
    int cnt = in[rt] = 0;
    rep(i, 0, n) f[i] = v[i] = -1;
    for (int i = 0, j; i <= n; i++)
    {
      res += in[i];
      for (j = i; j != rt&&v[j] != i&&f[j] == -1; v[j] = i, j = pre[j]);
      if (j != rt && f[j] == -1)
      {
        for (int k = pre[j]; k != j; k = pre[k]) f[k] = cnt;
        f[j] = cnt++;
      }
    }
    if (res - INF > INF - 2) return -1;
    if (!cnt) break;
    rep(i, 0, n) if (f[i] == -1) f[i] = cnt++;
    for (int i = 1, j; i <= m; i++)
    {
      x[i] = f[x[i]]; j = y[i]; y[i] = f[y[i]];
      if (x[i] != y[i]) z[i] -= in[j];
    }
    n = cnt - 1; rt = f[rt];
  }
  return res - INF + 1;
}

int main()
{
  while (intwo(n, m) != EOF)
  {
    rep(i, 1, m)
    {
      inthr(x[i], y[i], z[i]);
      x[i]++; y[i]++;
    }
    rep(i, 1, n) x[i + m] = 0, y[i + m] = i, z[i + m] = INF - 1;
    m += n; 
    int nn = n;
    if ((ans = zhuliu(0)) == -1) printf("impossible\n");
    else printf("%d %d\n", ans, who - m + nn - 1);
    putchar(10);
  }
  return 0;
}