BOJ 11376: 열혈강호 2

https://www.acmicpc.net/problem/11376

  • 2월동안 187문제 풀기(지난달 누적 87개) (32/187)
  • 열혈강호 1에서 유량만 바꾸면 맞는 문제.
  • 훨신 빠른 풀이도 있는듯 하다. Bimat 전용 풀이가 있는지 확인할 것.
  • Harpcroft-Karp 알고리즘 공부 필요
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#include <bits/stdc++.h>
using namespace std;

const int SZ = 1000;
int N, M, k, w, cap[2*SZ+2][2*SZ+2];
vector<int> adj[2*SZ+2];
const int S = 0, T = 2*SZ+1, INF = 0x7ffff;

void fillInput() {
  cin >> N >> M;
  for(int i = 1; i <= N; i++) {
    cin >> k;
    cap[S][i] = 1;
    adj[S].push_back(i);
    adj[i].push_back(S);
    for(int j = 0; j < k; j++) {
      cin >> w;
      w += SZ;
      adj[i].push_back(w);
      adj[w].push_back(i);
      cap[i][w] = 1;
    }
  }
  for(int i = SZ+1; i <= SZ + M; i++) {
    cap[i][T] = 1;
    adj[i].push_back(T);
    adj[T].push_back(i);
  }
}

int bfs(vector<int>& parent) {
  fill(parent.begin(), parent.end(), 0);
  queue<pair<int,int>> q;
  parent[S] = -1;
  q.push({S, INF});

  while(q.size()) {
    int cur = q.front().first;
    int flow = q.front().second;
    q.pop();

    for(int next : adj[cur]) {
      if(!parent[next] && cap[cur][next]) {
        parent[next] = cur;
        int newFlow = min(flow, cap[cur][next]);
        if (next == T) return newFlow;
        q.push({next, newFlow});
      }
    }
  }
  return 0;
}

void print() {
  for(auto row: cap) {
    for(int i = 0; i < 2*SZ+2; i++) cout << row[i] << " ";
    cout << "\n";
  }
  cout << "\n";
}

int maxflow() {
  int flow = 0;
  vector<int> parent(2*SZ+2);
  int newFlow = 0;

  //print();
  while (newFlow = bfs(parent)) {
    flow += newFlow;
    int cur = T;
    while (cur != S) {
      int prev = parent[cur];
      cap[prev][cur] -= newFlow;
      cap[cur][prev] += newFlow;
      cur = prev;
    }
    //print();
  }
  return flow;
}


int main() {
  fillInput();
  cout << maxflow();
}