二分图最大匹配模板(匈牙利算法)
P3386 【模板】二分图最大匹配 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)
struct augment_path {
vector<vector<int> > g;
vector<int> pa; // 匹配
vector<int> pb;
vector<int> vis; // 访问
int n, m; // 两个点集中的顶点数量
int dfn; // 时间戳记
int res; // 匹配数
augment_path(int _n, int _m) : n(_n), m(_m) {
assert(0 <= n && 0 <= m);
pa = vector<int>(n, -1);
pb = vector<int>(m, -1);
vis = vector<int>(n);
g.resize(n);
res = 0;
dfn = 0;
}
void add(int from, int to) {
assert(0 <= from && from < n && 0 <= to && to < m);
g[from].push_back(to);
}
bool dfs(int v) {
vis[v] = dfn;
for (int u : g[v]) {
if (pb[u] == -1) {
pb[u] = v;
pa[v] = u;
return true;
}
}
for (int u : g[v]) {
if (vis[pb[u]] != dfn && dfs(pb[u])) {
pa[v] = u;
pb[u] = v;
return true;
}
}
return false;
}
int solve() {
while (true) {
dfn++;
int cnt = 0;
for (int i = 0; i < n; i++) {
if (pa[i] == -1 && dfs(i)) {
cnt++;
}
}
if (cnt == 0) {
break;
}
res += cnt;
}
return res;
}
};