BOJ 11378: 열혈강호 4 작성일 2020-02-22 https://www.acmicpc.net/problem/11378 2월동안 187문제 풀기(지난달 누적 87개) (56/187) 이거 풀면 열혈강호 3도 공짜임 그래프 모델링, 그리고 Edmond-Karp로 안풀림. Dinic 필요 실상은 이게 유량문제인지 파악하는게 더 관건인듯. Dinic 이해는 했으나 코드 외워보기. 123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108#include <bits/stdc++.h> using namespace std; struct FlowEdge { int v, u; long long cap, flow = 0; FlowEdge(int v, int u, long long cap) : v(v), u(u), cap(cap) {} }; struct Dinic { const long long flow_inf = 1e18; vector<FlowEdge> edges; vector<vector<int>> adj; int n, m = 0; int s, t; vector<int> level, ptr; queue<int> q; Dinic(int n, int s, int t) : n(n), s(s), t(t) { adj.resize(n); level.resize(n); ptr.resize(n); } void add_edge(int v, int u, long long cap) { edges.emplace_back(v, u, cap); edges.emplace_back(u, v, 0); adj[v].push_back(m); adj[u].push_back(m + 1); m += 2; } bool bfs() { while (!q.empty()) { int v = q.front(); q.pop(); for (int id : adj[v]) { if (edges[id].cap - edges[id].flow < 1) continue; if (level[edges[id].u] != -1) continue; level[edges[id].u] = level[v] + 1; q.push(edges[id].u); } } return level[t] != -1; } long long dfs(int v, long long pushed) { if (pushed == 0) return 0; if (v == t) return pushed; for (int& cid = ptr[v]; cid < (int)adj[v].size(); cid++) { int id = adj[v][cid]; int u = edges[id].u; if (level[v] + 1 != level[u] || edges[id].cap - edges[id].flow < 1) continue; long long tr = dfs(u, min(pushed, edges[id].cap - edges[id].flow)); if (tr == 0) continue; edges[id].flow += tr; edges[id ^ 1].flow -= tr; return tr; } return 0; } long long flow() { long long f = 0; while (true) { fill(level.begin(), level.end(), -1); level[s] = 0; q.push(s); if (!bfs()) break; fill(ptr.begin(), ptr.end(), 0); while (long long pushed = dfs(s, flow_inf)) { f += pushed; } } return f; } }; int N, M, K, S, T, X; int main() { cin >> N >> M >> K; S = 0, T = N+M+1, X = N+M+2; Dinic dinic(N+M+3,S,T); dinic.add_edge(S, X, K); for(int i = 1; i <= N; i++) { dinic.add_edge(S, i, 1); dinic.add_edge(X, i, 1); // 열혈강호 4 (1 --> K) int c, w; cin >> c; for(int j = 0; j < c; j++) { cin >> w; w += N; dinic.add_edge(i, w, 1); } } for(int w = N+1; w <= N+M; w++) dinic.add_edge(w, T, 1); cout << dinic.flow(); }