HDU6214 Smallest Minimum Cut【最小割-最小边数】

Smallest Minimum Cut

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 471    Accepted Submission(s): 172

Problem Description

Consider a network G=(V,E) with source s and sink t. An s-t cut is a partition of nodes set V into two parts such that s and t belong to different parts. The cut set is the subset of E with all edges connecting nodes in different parts. A minimum cut is the one whose cut set has the minimum summation of capacities. The size of a cut is the number of edges in the cut set. Please calculate the smallest size of all minimum cuts.

 

Input

The input contains several test cases and the first line is the total number of casesT (1≤T≤300).
Each case describes a network G, and the first line contains two integers n (2≤n≤200) and m (0≤m≤1000) indicating the sizes of nodes and edges. All nodes in the network are labelled from1 to n.
The second line contains two different integers s and t (1≤s,t≤n) corresponding to the source and sink.
Each of the next m lines contains three integers u,v and w (1≤w≤255) describing a directed edge from node u to v with capacity w.

 

Output

For each test case, output the smallest size of all minimum cuts in a line.

 

Sample Input

24 51 41 2 31 3 12 3 12 4 13 4 24 51 41 2 31 3 12 3 12 4 13 4 3

 

Sample Output

23

 

 题意：给点一个有向图（n个点（n<=200）m条边（m<=1000）），求从s到t的最小割的最小边数。

思路：将每条边权值扩大大数MOD倍再加上1，所求最大流膜MOD即为答案。跑完最大流，假设在乘大数不加一的情况下，边容量为0即为割边，加1后，多跑的最大流即为最小割边的最小边数。（很奇怪，我用另外一种，先跑完最大流，然后根据容量新建边（cap为0建1，否则为INF）一直wa）

ACcode（Dinic）

#include<bits/stdc++.h>using namespace std;const int N=1000;const int MOD=100000;const int INF=0x3f3f3f3f;struct edge{    int to,cap,rev;    edge(int x,int y,int z)    {        to=x;cap=y;rev=z;    }};vector<edge> g[N];int level[N],iter[N],n,m,s,t;bool vis[N];void add(int from,int to,int cap){    g[from].push_back(edge(to,cap,g[to].size()));    g[to].push_back(edge(from,0,g[from].size()-1));}void bfs(int s){    memset(level,-1,sizeof(level));    level[s]=0;    queue<int> que;    que.push(s);    while(!que.empty())    {        int v=que.front();que.pop();        for(int i=0;i<g[v].size();i++)        {            edge &e=g[v][i];            if(e.cap>0&&level[e.to]<0)            {                level[e.to]=level[v]+1;                que.push(e.to);            }        }    }}int dfs(int v,int t,int f){    if(v==t)return f;    for(int &i=iter[v];i<g[v].size();i++)    {        edge &e=g[v][i];        if(e.cap>0&&level[e.to]>level[v])        {            int d=dfs(e.to,t,min(f,e.cap));            if(d>0)            {                e.cap-=d;                g[e.to][e.rev].cap+=d;                return d;            }        }    }    return 0;}int max_flow(int s,int t){    int  flow=0;    while(1)    {        bfs(s);        if(level[t]<0)return flow;        memset(iter,0,sizeof(iter));        int f;        while(f=dfs(s,t,INF))        {            flow+=f;        }    }}int main(){    int TA,x,y,z;    scanf("%d",&TA);    while(TA--)    {        scanf("%d%d%d%d",&n,&m,&s,&t);        for(int i=1;i<=n;i++)            g[i].clear();        for(int i=1;i<=m;i++)        {            scanf("%d%d%d",&x,&y,&z);            z=z*MOD+1;            add(x,y,z);        }        printf("%d\n",max_flow(s,t)%MOD);    }    return 0;}