Smallest Minimum Cut HDU

Smallest Minimum Cut

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2281    Accepted Submission(s): 913

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 cases T (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 from 1 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

2 4 5 1 4 1 2 3 1 3 1 2 3 1 2 4 1 3 4 2 4 5 1 4 1 2 3 1 3 1 2 3 1 2 4 1 3 4 3

 

Sample Output

2 3

 

就是求最小割的边的集合

 

最小割的边就等于满流的边

1、将每条边的权值改为w*(m+1)+1, 最后求的最大流除以(m+1)就是原图的最大流，模上(m+1)就是最小割的边

2、求得最大流之后，将所有的满流的边权设为1，不满流的边权设为INF，然后跑一边最大流就是最小割的边

 

 

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 10010, INF = 0x7fffffff;
int head[maxn], cnt;
int d[maxn], vis[maxn], cur[maxn];
int s, t;
struct node
{
    int u, v, c, next;
}Node[maxn<<1];

void add_(int u, int v, int c)
{
    Node[cnt].u = u;
    Node[cnt].v = v;
    Node[cnt].c = c;
    Node[cnt].next = head[u];
    head[u] = cnt++;
}

void add(int u, int v, int c)
{
    add_(u, v, c);
    add_(v, u, 0);
}



void init()
{
    mem(head, -1);
    cnt = 0;
}

bool bfs()
{
    queue<int> Q;
    mem(d, 0);
    Q.push(s);
    d[s] = 1;
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        for(int i=head[u]; i!=-1; i=Node[i].next)
        {
            node e = Node[i];
            if(!d[e.v] && e.c > 0)
            {
                d[e.v] = d[e.u] + 1;
                Q.push(e.v);
                if(e.v == t) return 1;
            }
        }
    }
    return d[t] != 0;
}

int dfs(int u, int cap)
{
    int ret = 0, V;
    if(u==t || cap == 0)
        return cap;
    for(int &i=cur[u]; i!=-1; i=Node[i].next)
    {
        node e = Node[i];
        if(d[e.v] == d[e.u] + 1 && e.c > 0)
        {
            int V = dfs(e.v, min(cap, e.c));
            Node[i].c -= V;
            Node[i^1].c += V;
            ret += V;
            cap -= V;
            if(cap == 0) break;
        }
    }
    if(cap > 0) d[u] = -1;
    return ret;
}

int dinic(int u)
{
    int ans = 0;
    while(bfs())
    {
        memcpy(cur, head, sizeof(head));
        ans += dfs(u, INF);
    }
    return ans;
}

int main()
{
    int T, n, m;
    rd(T);
    while(T--)
    {
        int u, v, c;
        init();
        rd(n); rd(m);
        rd(s), rd(t);
        rep(i, 0, m)
        {
            rd(u), rd(v), rd(c);
            add(u, v, c*(m+1) + 1);
        }

        int res = dinic(s);
        cout<< res % (m+1) <<endl;

    }

    return 0;
}