The regime of a small but wealthy dictatorship has been abruptly overthrown by an unexpected rebellion.
Because of the enormous disturbances this is causing in world economy, an imperialist military
super power has decided to invade the country and reinstall the old regime.
For this operation to be successful, communication between the capital and the largest city must
be completely cut. This is a difficult task, since all cities in the country are connected by a computer
network using the Internet Protocol, which allows messages to take any path through the network.
Because of this, the network must be completely split in two parts, with the capital in one part and
the largest city in the other, and with no connections between the parts.
There are large differences in the costs of sabotaging different connections, since some are much
more easy to get to than others.
Write a program that, given a network specification and the costs of sabotaging each connection,
determines which connections to cut in order to separate the capital and the largest city to the lowest
possible cost.
Input
Input file contains several sets of input. The description of each set is given below.
The first line of each set has two integers, separated by a space: First one the number of cities, n in
the network, which is at most 50. The second one is the total number of connections, m, at most 500.
The following m lines specify the connections. Each line has three parts separated by spaces: The
first two are the cities tied together by that connection (numbers in the range 1 − n). Then follows the
cost of cutting the connection (an integer in the range 1 to 40000000). Each pair of cites can appear
at most once in this list.
Input is terminated by a case where values of n and m are zero. This case should not be processed.
For every input set the capital is city number 1, and the largest city is number 2.
Output
For each set of input you should produce several lines of output. The description of output for each set
of input is given below:
The output for each set should be the pairs of cities (i.e. numbers) between which the connection
should be cut (in any order), each pair on one line with the numbers separated by a space. If there is
more than one solution, any one of them will do.
Print a blank line after the output for each set of input.
Sample Input
5 8
1 4 30
1 3 70
5 3 20
4 3 5
4 5 15
5 2 10
3 2 25
2 4 50
5 8
1 4 30
1 3 70
5 3 20
4 3 5
4 5 15
5 2 10
3 2 25
2 4 50
0 0
Sample Output
4 1
3 4
3 5
3 2
4 1
3 4
3 5
3 2

題意：一個國家出了內亂，現在想要把首都和最大的城市的通訊完全斷開，但是斷開一條邊有花費，現在問斷開哪些邊可以使花費最小。

分析：跑一遍最大流dinic後，根據最小割定理，整部圖被分為S,T兩部分，從源點出發dfs得到S集的點，剩下的為T集的點，然後輸出左右兩邊各一個連著的點就是割邊。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int inf = 0x3f3f3f3f;
const int mod=609929123;
const int N=100220;
int n,f,d,m;
int S,T,R;//源點，匯點
struct node
{
    int v,cap,next;//領接表
}es[4*N];
int frist[N];
int dis[N];
int current[N];//當前弧優化
int x[N],y[N];
int vis[N];
int addedge(int u,int v,int cap)
{
    node  e1={v,cap,frist[u]};
    es[R]=e1;
    frist[u]=R++;
}
int bfs()
{
  queue<int>q;
  q.push(S);
  memset(dis,-1,sizeof(dis));
  dis[S]=0;
  while(!q.empty())
  {
      int h=q.front();
      if(h==T) return 1;
      q.pop();
      for(int i=frist[h];i!=-1;i=es[i].next)
      {
          int temp=es[i].v;
          if(dis[temp]==-1&&es[i].cap)
          {
              dis[temp]=dis[h]+1;
              q.push(temp);
          }
      }
  }
  return 0;
}
int dinic(int x,int maxflow)
{
  if(x==T) return maxflow;
  int flow,f=0;
  for(int &i=current[x];i!=-1;i=es[i].next)
  {
      int temp=es[i].v;
      if(dis[temp]==dis[x]+1&&es[i].cap)
      {
      flow=dinic(temp,min(maxflow-f,es[i].cap));
      es[i].cap-=flow;
      es[i^1].cap+=flow;
      f+=flow;
      if(f==maxflow) break;
      }
  }
  return f;
}
int DINIC()
{
 int ans=0;
 while(bfs())
 {
     int flow;
     memcpy(current,&frist,sizeof(frist));
     while(flow=dinic(S,inf))
     ans+=flow;
 }
 return ans;
}
void dfs(int x)//最小割 找出s部分的點
{
    vis[x]=1;
    for(int i=frist[x];i!=-1;i=es[i].next)
    {
        int temp=es[i].v;
        if(!vis[temp]&&es[i].cap)
             dfs(temp);
    }
}
int main()
{
    int u,v,cap;
   while(~scanf("%d%d",&n,&m)&&(n+m))
   {
       memset(frist,-1,sizeof(frist));
       R=0,S=1,T=2;
       for(int i=1;i<=m;i++)
       {
           scanf("%d%d%d",&x[i],&y[i],&cap);
           addedge(x[i],y[i],cap);
           addedge(y[i],x[i],cap);
       }
      DINIC();
      memset(vis,0,sizeof(vis));
      dfs(S);
      for(int i=1;i<=m;i++)
      {
          if((vis[x[i]]&&!vis[y[i]])||(vis[y[i]]&&!vis[x[i]]))
            printf("%d %d\n",x[i],y[i]);
      }
      printf("\n");
   }
}