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 }
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