题意
给出一张有向无环图,求出用最少的路径覆盖整张图,要求路径在定点处不相交
输出方案
Sol
定理:路径覆盖 = 定点数 - 二分图最大匹配数
直接上匈牙利
输出方案的话就不断的从一个点跳匹配边
#include<cstdio> #include<queue> #include<cstring> using namespace std; const int MAXN = 1e5 + 10, INF = 1e9 + 10; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } int N, M; vector<int> v[MAXN]; int link[MAXN], vis[MAXN], cnt = 1; bool Arg(int x) { for(int i = 0; i < v[x].size(); i++) { int to = v[x][i]; if(vis[to] == cnt) continue; vis[to] = cnt; if(!link[to] || Arg(link[to])) {link[to] = x; link[x] = to; return 1;} } return 0; } int Hunary() { int ans = 0; for(int i = 1; i <= N; i++, cnt++) if(Arg(i)) ans++; return ans; } int main() { N = read(); M = read(); for(int i = 1; i <= M; i++) { int x = read(), y = read(); v[x].push_back(y + N); } int ans = N - Hunary(); memset(vis, 0, sizeof(vis)); for(int i = 1; i <= N; i++) { int x = i + N; if(vis[i]) continue; do printf("%d ", x = x - N); while(vis[x] = 1, x = link[x]); puts(""); } printf("%d", ans); return 0; }