기본적으로 백준 7616번과 완전하게 동일하다.
(viyoung.tistory.com/159)
2번이상 방문하지 않는다는 상황을 정점 분할해서 처리해주면 된다.
정점 분할하는 방법은 해당 index에 2배 처리해줘서 찢어주면 된다.
#include <bits/stdc++.h>
#define fastio ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#define INF 123456789;
const int MAX_N = 802;
using namespace std;
struct edge{
int from, to, capacity, flow;
edge* reverse_edge;
edge(int u, int v, int c) : from(u), to(v), capacity(c), flow(0)
{}
int residual(){
return capacity - flow;
}
void add_flow(int amount){
flow += amount;
reverse_edge -> flow -= amount;
}
};
vector<edge*> adj[MAX_N];
void add_edge(int u, int v, int c, bool dir = true){
edge* e1 = new edge(u, v, c);
edge* e2 = new edge(v, u, dir ? 0 : c);
e1 -> reverse_edge = e2;
e2 -> reverse_edge = e1;
adj[u].push_back(e1);
adj[v].push_back(e2);
}
int networkFlow(int source, int sink){
int max_flow = 0;
while(true){
vector<edge*> parent(MAX_N, nullptr);
queue<int> q;
q.push(source);
while(!q.empty() && parent[sink] == nullptr){
int here = q.front(); q.pop();
for(int i = 0; i < adj[here].size(); i++){
if(adj[here][i] -> residual() > 0 && parent[adj[here][i] -> to] == nullptr){
q.push(adj[here][i] -> to);
parent[adj[here][i] -> to] = adj[here][i];
}
}
}
if(parent[sink] == nullptr) break; // There is no augmenting path
// 무조건 minimum capacity가 1이므로 해당 작업은 생략해도 무방하다.
for(int p = sink; p != source; p = parent[p] -> from){
if(p != 1 && p != 3)parent[p] -> add_flow(1);
}
max_flow++;
}
return max_flow;
}
int main(void){
fastio;
int N, P;
cin >> N >> P;
for(int i = 1; i <= N; i++){
add_edge(i * 2 - 2, i * 2 - 1, 1);
}
for(int i = 0; i < P; i++){
int num1, num2;
cin >> num1 >> num2;
add_edge(num1 * 2 - 1, num2 * 2 - 2, 1);
add_edge(num2 * 2 - 1, num1 * 2 - 2, 1);
}
cout << networkFlow(0, 3) << "\n";
int n;
return 0;
}