Smallest Minimum Cut
Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2420 Accepted Submission(s): 973
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
Source
2017 ACM/ICPC Asia Regional Qingdao Online
在所有最小割中找出边的数目最小的，和HDU3987一样，添加边的时候流量改为w*（m + 1） + 1，跑一遍最大流Max，最小割为Max/（m+1），最少割边数为Max%（m+1）
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 3005;
const int INF = 0x3f3f3f3f;
struct Edge1
{
int from,to,cap,flow;
};
struct Dinic
{
int n,m,s,t;
vector<Edge1> edges;
vector<int> G[MAXN];
bool vis[MAXN];
int d[MAXN];
int cur[MAXN];
void init(int n)
{
this -> n = n;
for(int i = 0; i <= n + 1; i++){
G[i].clear();
}
edges.clear();
}
void AddEdge(int from,int to,int cap)
{
edges.push_back((Edge1){from,to,cap,0});
edges.push_back((Edge1){to,from,0,0});
m = edges.size();
G[from].push_back(m - 2);
G[to].push_back(m - 1);
}
bool BFS()
{
memset(vis,0,sizeof(vis));
queue<int> Q;
Q.push(s);
d[s] = 0;
vis[s] = 1;
while(!Q.empty()) {
int x = Q.front();
Q.pop();
for(int i = 0; i < G[x].size(); i++) {
Edge1& e = edges[G[x][i]];
if(!vis[e.to] && e.cap > e.flow) {
vis[e.to] = 1;
d[e.to] = d[x] + 1;
Q.push(e.to);
}
}
}
return vis[t];
}
int DFS(int x,int a)
{
if(x == t || a == 0) return a;
int flow = 0,f;
for(int& i = cur[x]; i < G[x].size(); i++) {
Edge1& e = edges[G[x][i]];
if(d[x] + 1 == d[e.to] && (f = DFS(e.to,min(a,e.cap - e.flow))) > 0) {
e.flow += f;
edges[G[x][i] ^ 1].flow -= f;
flow += f;
a -= f;
if(a == 0) break;
}
}
return flow;
}
int Maxflow(int s,int t) {
this -> s = s,this -> t = t;
int flow = 0;
while(BFS()) {
memset(cur,0,sizeof(cur));
flow += DFS(s,INF);
}
return flow;
}
}din;
int main(void)
{
int T,n,m,s,t,u,v,w,ans;
scanf("%d",&T);
while(T--) {
scanf("%d %d",&n,&m);
scanf("%d %d",&s,&t);
din.init(n + 2);
for(int i = 1; i <= m; i++) {
scanf("%d %d %d",&u,&v,&w);
din.AddEdge(u,v,w * (m + 1) + 1);
}
ans = din.Maxflow(s,t);
printf("%d\n",ans % (m + 1));
}
return 0;
}