UVA-10480 Sabotage（最大流最小割定理+输出路径）

题目链接：UVA-10480 Sabotage

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

题意：

含有n个结点及m条无向边的图，每条边都带有边权（一个正整数），现要删除边使得结点s（编号为1）和t（编号为2）不连通，要令删除的边权之和最小，要求输出被删除的边（以任意顺序输出任意一个满足的方案）

分析：

这是一个最小割问题，由最大流最小割定理可知：s-t 最小割的容量等于 s-t 最大流的流量

那么求最小割就转化为了求最大流，但是要注意，求最小割时是无向图，所以正向边和反向边的初始残量都要等于边权

这题有一个难点，并不是要求输出最小割，而是输出达到最小割时需要割掉的边。

因为跑了一遍最大流后，图会被分成两部分（结点s所在的部分和结点t所在的部分），所以只需要先预处理一下哪些点所在部分不同，然后遍历所有边，起点、终点所在部分不同的边即是需要割掉的边。（为方便处理，建图时要记录边的起点和终点）

以下代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<queue>
#include<map>
#include<algorithm>
#define LL long long
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=1e2+50;
const int maxm=1e3+50;
struct edge
{int w;int from;int to;int next;
}e[maxm];
int head[maxn],cnt;
int s,t;
int n,m;
void addedge(int u,int v,int w)
{e[cnt].from=u;e[cnt].w=w;e[cnt].to=v;e[cnt].next=head[u];head[u]=cnt;cnt++;
}
int dis[maxn];     //dis数组记录层次
bool bfs()         //利用BFS建立分成图，从而可以多次DFS增广
{memset(dis,-1,sizeof(dis));     //初始化dis数组queue<int> q;q.push(s);dis[s]=0;      //源点层次为0while(!q.empty()){int u=q.front();q.pop();for(int i=head[u];i!=-1;i=e[i].next){int v=e[i].to;if(e[i].w>0&&dis[v]==-1)     //可达&&未分层{dis[v]=dis[u]+1;         //分层if(v==t)                 //若到达汇点，则分层结束，返回truereturn true;q.push(v);}}}return false;  //运行到此处，说明汇点已不可达，返回false
}
int cur[maxn];     //弧优化：cur数组用于记录上一次DFS增广时u已经增广到第几条边，从而优化时间
int dfs(int u,int flow)       //flow代表流入u点的最大流量
{if(u==t)return flow;          //到达汇点，直接返回flowfor(int &i=cur[u];i!=-1;i=e[i].next){                         //注意i前面用&引用，这样就可以直接改变cur[u]int v=e[i].to;if(dis[v]==dis[u]+1&&e[i].w>0)   //v为u的下一层&&可达{int k=dfs(v,min(flow,e[i].w));if(k>0){e[i].w-=k;         //正向边-=ke[i^1].w+=k;       //反向边+=kreturn k;}}}return 0;       //无法继续增广，返回0
}int dinic()
{int ans=0;      //记录总流量while(bfs())    //分层{for(int i=0;i<maxn;i++)    //初始化cur数组，即将head数组赋给cur数组cur[i]=head[i];while(int k=dfs(s,INF))    //增广ans+=k;}return ans;
}
bool vis[maxn];
void dfs_again(int u)
{if(vis[u])return;elsevis[u]=true;for(int i=head[u];i!=-1;i=e[i].next){if(e[i].w>0)        //残量>0才能通过dfs_again(e[i].to);}
}
int main()
{while(scanf("%d %d",&n,&m)&&n+m){memset(head,-1,sizeof(head));cnt=0;s=1;t=2;for(int i=1;i<=m;i++){int u,v,w;scanf("%d %d %d",&u,&v,&w);addedge(u,v,w);addedge(v,u,w);}dinic();memset(vis,false,sizeof(vis));dfs_again(s);            //遍历一遍s所在部分，加上标记bool flag[maxm]={false};for(int i=0;i<cnt;i++)   //遍历所有边{if(flag[i])continue;int u=e[i].from,v=e[i].to;if((!vis[u]&&vis[v])||(!vis[v]&&vis[u])){printf("%d %d\n",u,v);flag[i]=true;flag[i^1]=true;  //防止多输出一次反向边}}printf("\n");}return 0;
}

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UVA-10480 Sabotage（最大流最小割定理+输出路径）

题意：

分析：

以下代码：

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