这几天刷最大流和最小割，正好某谷推荐了这道题，就顺便AC了一下。
好水。
建立源点和汇点，对于每一个能提供床位的同学，就从源点向该同学连一条容量为1的边；对于每一个需要床位的同学，就向汇点连一条容量为1的边。因此将每一个同学拆点。
接下来考虑同学之间的关系：如果x和y互相认识，就连一条(x, y +n)，(x +n, y)容量为1的边。
然后跑最大流，看总流量是否等于需要床位的人数。
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, x) memset(a, x, sizeof(a)) 15 #define rg register 16 typedef long long ll; 17 typedef double db; 18 const int INF = 0x3f3f3f3f; 19 const db eps = 1e-8; 20 const int maxn = 55; 21 inline ll read() 22 { 23 ll ans = 0; 24 char ch = getchar(), last = ' '; 25 while(!isdigit(ch)) {last = ch; ch = getchar();} 26 while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();} 27 if(last == '-') ans = -ans; 28 return ans; 29 } 30 inline void write(ll x) 31 { 32 if(x < 0) x = -x, putchar('-'); 33 if(x >= 10) write(x / 10); 34 putchar(x % 10 + '0'); 35 } 36 37 int n, t, tot = 0; 38 bool in[maxn]; 39 40 struct Edge 41 { 42 int from, to, cap, flow; 43 }; 44 vector<Edge> edges; 45 vector<int> G[maxn << 1]; 46 void addEdge(int from, int to) 47 { 48 edges.push_back((Edge){from, to, 1, 0}); 49 edges.push_back((Edge){to, from, 0, 0}); 50 int sz = edges.size(); 51 G[from].push_back(sz - 2); 52 G[to].push_back(sz - 1); 53 } 54 55 int dis[maxn << 1]; 56 bool bfs() 57 { 58 Mem(dis, 0); dis[0] = 1; 59 queue<int> q; q.push(0); 60 while(!q.empty()) 61 { 62 int now = q.front(); q.pop(); 63 for(int i = 0; i < (int)G[now].size(); ++i) 64 { 65 Edge& e = edges[G[now][i]]; 66 if(!dis[e.to] && e.cap > e.flow) 67 { 68 dis[e.to] = dis[now] + 1; 69 q.push(e.to); 70 } 71 } 72 } 73 return dis[t]; 74 } 75 int cur[maxn << 1]; 76 int dfs(int now, int res) 77 { 78 if(now == t || res == 0) return res; 79 int flow = 0, f; 80 for(int& i = cur[now]; i < (int)G[now].size(); ++i) 81 { 82 Edge& e = edges[G[now][i]]; 83 if(dis[e.to] == dis[now] + 1 && (f = dfs(e.to, min(res, e.cap - e.flow))) > 0) 84 { 85 e.flow += f; 86 edges[G[now][i] ^ 1].flow -= f; 87 flow += f; res -= f; 88 if(res == 0) break; 89 } 90 } 91 return flow; 92 } 93 94 int maxflow() 95 { 96 int flow = 0; 97 while(bfs()) 98 { 99 Mem(cur, 0); 100 flow += dfs(0, INF); 101 } 102 return flow; 103 } 104 105 void init(int n) 106 { 107 edges.clear(); 108 for(int i = 0; i <= n; ++i) G[i].clear(); 109 Mem(in, 0); 110 tot = 0; 111 } 112 113 int main() 114 { 115 int T = read(); 116 while(T--) 117 { 118 n = read(); t = (n << 1) + 1; 119 init(t); 120 for(int i = 1; i <= n; ++i) 121 { 122 in[i] = (bool)read(); 123 if(in[i]) addEdge(0, i); 124 else addEdge(i + n, t), tot++; 125 } 126 for(int i = 1; i <= n; ++i) 127 { 128 int x = read(); 129 if(in[i] && !x) addEdge(i + n, t), tot++; 130 } 131 for(int i = 1; i <= n; ++i) 132 for(int j = 1; j <= n; ++j) 133 { 134 int x = read(); 135 if(x || i == j) addEdge(i, j + n), addEdge(j, i + n); 136 } 137 printf("%s ", maxflow() == tot ? "^_^" : "T_T"); 138 } 139 return 0; 140 }