题解：如果一个人一条边和适应星球连一条流量为1的边，那么结果因为点数太多，边数太多而TLE。现在我们把所有的适应星球状态一样的人放在一起，然后在同一个状态的人的边一起连边，从而减少边数和点数。
代码：
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); 4 #define LL long long 5 #define ULL unsigned LL 6 #define fi first 7 #define se second 8 #define pb emplace_back 9 #define lson l,m,rt<<1 10 #define rson m+1,r,rt<<1|1 11 #define lch(x) tr[x].son[0] 12 #define rch(x) tr[x].son[1] 13 #define max3(a,b,c) max(a,max(b,c)) 14 #define min3(a,b,c) min(a,min(b,c)) 15 typedef pair<int,int> pll; 16 const int inf = 0x3f3f3f3f; 17 const LL INF = 0x3f3f3f3f3f3f3f3f; 18 const LL mod = (int)1e9+7; 19 const int N = 1024 + 100; 20 const int M = N * 100; 21 int head[N], deep[N], cur[N]; 22 int w[M], to[M], nx[M]; 23 int tot; 24 void add(int u, int v, int val){ 25 w[tot] = val; to[tot] = v; 26 nx[tot] = head[u]; head[u] = tot++; 27 28 w[tot] = 0; to[tot] = u; 29 nx[tot] = head[v]; head[v] = tot++; 30 } 31 int bfs(int s, int t){ 32 queue<int> q; 33 memset(deep, 0, sizeof(deep)); 34 q.push(s); 35 deep[s] = 1; 36 while(!q.empty()){ 37 int u = q.front(); 38 q.pop(); 39 for(int i = head[u]; ~i; i = nx[i]){ 40 if(w[i] > 0 && deep[to[i]] == 0){ 41 deep[to[i]] = deep[u] + 1; 42 q.push(to[i]); 43 } 44 } 45 } 46 return deep[t] > 0; 47 } 48 int Dfs(int u, int t, int flow){ 49 if(u == t) return flow; 50 for(int &i = cur[u]; ~i; i = nx[i]){ 51 if(deep[u]+1 == deep[to[i]] && w[i] > 0){ 52 int di = Dfs(to[i], t, min(w[i], flow)); 53 if(di > 0){ 54 w[i] -= di, w[i^1] += di; 55 return di; 56 } 57 } 58 } 59 return 0; 60 } 61 62 int Dinic(int s, int t){ 63 int ans = 0, tmp; 64 while(bfs(s, t)){ 65 for(int i = 0; i <= t; i++) cur[i] = head[i]; 66 while(tmp = Dfs(s, t, inf)) ans += tmp; 67 } 68 return ans; 69 } 70 int a[N]; 71 void init(){ 72 memset(head, -1, sizeof(head)); 73 memset(a, 0, sizeof(a)); 74 tot = 0; 75 } 76 int main(){ 77 int n, m; 78 while(~scanf("%d%d", &n, &m)){ 79 init(); 80 int Max = (1 << m)- 1, val; 81 for(int i = 1; i <= n; ++i){ 82 int tmp = 0; 83 for(int i = 1; i <= m; ++i){ 84 scanf("%d", &val); 85 tmp = tmp * 2 + val; 86 } 87 ++a[tmp]; 88 } 89 int s = 0, t = Max + m + 1; 90 for(int i = 1; i <= m; ++i){ 91 scanf("%d", &val); 92 add(i + Max, t, val); 93 } 94 for(int i = 1; i <= Max; ++i){ 95 add(s, i, a[i]); 96 for(int j = 1; j <= m; ++j){ 97 if((i>>(m-j))&1) { 98 add(i, j+Max, inf); 99 } 100 } 101 } 102 if(Dinic(s,t) == n) puts("YES"); 103 else puts("NO"); 104 } 105 return 0; 106 }