这里就讲怎么做……因为为什么这么做以及证明我都不知道……
首先，我们将原图的每一个点 i 都拆成 i 和 i +n 两个点。接着把所有 i 都和源点相连，边的容量为1,；把所有i + n 都和汇点相连，容量也为1。然后对于原图中的一条边(u, v)，就在新图中连一条(u, v + n)的边。对该图跑最大流，则答案就是原来节点数 n - maxflow().
对于输出每个覆盖，我的方法比较暴力：在跑完maxflow()的残图上跑，如果该边流满了，就递归找这个节点的所有边。
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<cstdlib> 7 #include<cctype> 8 #include<vector> 9 #include<stack> 10 #include<queue> 11 using namespace std; 12 #define enter puts("") 13 #define space putchar(' ') 14 #define Mem(a) memset(a, 0, sizeof(a)) 15 typedef long long ll; 16 typedef double db; 17 const int INF = 0x3f3f3f3f; 18 const db eps = 1e-8; 19 const int maxn = 155; 20 inline ll read() 21 { 22 ll ans = 0; 23 char ch = getchar(), last = ' '; 24 while(!isdigit(ch)) {last = ch; ch = getchar();} 25 while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();} 26 if(last == '-') ans = -ans; 27 return ans; 28 } 29 inline void write(ll x) 30 { 31 if(x < 0) x = -x, putchar('-'); 32 if(x >= 10) write(x / 10); 33 putchar(x % 10 + '0'); 34 } 35 36 int n, m, t; 37 struct Edge 38 { 39 int from, to, cap, flow; 40 }; 41 vector<Edge> edges; 42 vector<int> G[maxn << 1]; 43 void addEdge(int from, int to) 44 { 45 edges.push_back((Edge){from, to, 1, 0}); 46 edges.push_back((Edge){to, from, 0, 0}); 47 int sz = edges.size(); 48 G[from].push_back(sz - 2); 49 G[to].push_back(sz - 1); 50 } 51 52 int dis[maxn << 1]; 53 bool bfs() 54 { 55 Mem(dis); dis[0] = 1; 56 queue<int> q; q.push(0); 57 while(!q.empty()) 58 { 59 int now = q.front(); q.pop(); 60 for(int i = 0; i < (int)G[now].size(); ++i) 61 { 62 Edge& e = edges[G[now][i]]; 63 if(!dis[e.to] && e.cap > e.flow) 64 { 65 dis[e.to] = dis[now] + 1; 66 q.push(e.to); 67 } 68 } 69 } 70 return dis[t]; 71 } 72 int cur[maxn << 1]; 73 int dfs(int now, int res) 74 { 75 if(now == t || res == 0) return res; 76 int flow = 0, f; 77 for(int& i = cur[now]; i < (int)G[now].size(); ++i) 78 { 79 Edge& e = edges[G[now][i]]; 80 if(dis[e.to] == dis[now] + 1 && (f = dfs(e.to, min(res, e.cap - e.flow))) > 0) 81 { 82 e.flow += f; 83 edges[G[now][i] ^ 1].flow -= f; 84 flow += f; 85 res -= f; 86 if(res == 0) break; 87 } 88 } 89 return flow; 90 } 91 92 int maxflow() 93 { 94 int flow = 0; 95 while(bfs()) 96 { 97 Mem(cur); 98 flow += dfs(0, INF); 99 } 100 return flow; 101 } 102 103 bool vis[maxn]; 104 void print(int now) 105 { 106 if(now <= 0) return; 107 vis[now] = 1; 108 write(now); space; 109 for(int i = 0; i < (int)G[now].size(); ++i) 110 { 111 Edge& e = edges[G[now][i]]; 112 if(e.cap == e.flow) {print(e.to - n); break;} 113 } 114 } 115 116 int main() 117 { 118 n = read(); m = read(); 119 t = n + n + 1; 120 for(int i = 1; i <= n; ++i) addEdge(0, i), addEdge(i + n, t); 121 for(int i = 1; i <= m; ++i) 122 { 123 int x = read(), y = read(); 124 addEdge(x, y + n); 125 } 126 int ans = n - maxflow(); 127 for(int i = 1; i <= n; ++i) if(!vis[i]) 128 print(i), enter; 129 write(ans); enter; 130 return 0; 131 }