这道题其实蛮好想的，因为分为正，副飞行员。所以就把正飞行员当作Boy，副飞行员当作Girl。然后做Hungry即可。
1 #include<bits/stdc++.h>
2 using namespace std;
3 int G[131][131],vis[131],girl[131],n,n1;
4 int hungry(int boy)
5 {
6 int i;
7 for(i=n1+1;i<=n;i++)
8 {
9 if(G[boy][i]==1&&vis[i]==0)
10 {
11 vis[i]=1;
12 if(hungry(girl[i])==1||girl[i]==0)
13 {
14 girl[i]=boy;
15 return 1;
16 }
17 }
18 }
19 return 0;
20 }
21 int main()
22 {
23 int a,b,sum,i;
24 scanf("%d %d",&n,&n1);
25 while(scanf("%d %d",&a,&b)!=EOF)G[a][b]=1;
26 memset(girl,0,sizeof(girl));
27 sum=0;
28 for(i=1;i<=n1;i++)
29 {
30 memset(vis,0,sizeof(vis));
31 if(hungry(i)==1)sum++;
32 }
33 printf("%d",sum);
34 return 0;
35 }
但这道题是“网络流24题”，所以还是要练一练网络流。所以直接用Dinic做最大流即可。
1 #include<cstdio>
2 #include<cstring>
3 #include<cmath>
4 #include<cstdlib>
5 #include<algorithm>
6 using namespace std;
7 int n,s,t,q[131],dis[131],G[131][131];
8 int Min(int aaa,int bbb){if(aaa>bbb)return bbb;else return aaa;}
9 int Bfs()
10 {
11 int head,tail,u,v;
12 memset(q,0,sizeof(q)),head=0,tail=1,q[tail]=s;
13 memset(dis,-1,sizeof(dis)),dis[s]=0;
14 while(head<=tail)
15 {
16 u=q[++head];
17 for(v=1;v<=n;v++)
18 {
19 if(G[u][v]>0&&dis[v]<0)
20 {
21 dis[v]=dis[u]+1;
22 q[++tail]=v;
23 }
24 }
25 }
26 if(dis[n]<=0)return 0;
27 else return 1;
28 }
29 int Dfs(int u,int minflow)
30 {
31 int v,ans;
32 if(u==n)return minflow;
33 for(v=1;v<=n;v++)
34 {
35 if(G[u][v]>0&&dis[v]==dis[u]+1)
36 {
37 ans=Dfs(v,Min(minflow,G[u][v]));
38 if(ans!=0)
39 {
40 G[u][v]-=ans;
41 G[v][u]+=ans;
42 return ans;
43 }
44 }
45 }
46 return 0;
47 }
48 int Dinic()
49 {
50 int maxflow=0,ans;
51 while(Bfs()!=0)
52 {
53 ans=Dfs(s,0x7fffffff);if(ans==0)break;maxflow+=ans;
54 }
55 return maxflow;
56 }
57 int main()
58 {
59 int n1,a,b,i;
60 scanf("%d %d",&n,&n1);
61 while(scanf("%d %d",&a,&b)!=EOF)G[a][b]=1;
62 n+=2;s=n-1;t=n;
63 for(i=1;i<=n1;i++)G[s][i]=1;
64 for(i=n1+1;i<=n-2;i++)G[i][t]=1;
65 printf("%d",Dinic());
66 return 0;
67 }