Cogs 14. [网络流24题] 搭配飞行员

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。

 #include<bits/stdc++.h>
 using namespace std;
 int G[][],vis[],girl[],n,n1;
 int hungry(int boy)
 {
     int i;
     for(i=n1+;i<=n;i++)
     {
         if(G[boy][i]==&&vis[i]==)
         {
             vis[i]=;
             if(hungry(girl[i])==||girl[i]==)
             {
                 girl[i]=boy;
                 return ;
             }
         }
     }
     return ;
 }
 int main()
 {
     int a,b,sum,i;
     scanf("%d %d",&n,&n1);
     while(scanf("%d %d",&a,&b)!=EOF)G[a][b]=;
     memset(girl,,sizeof(girl));
     sum=;
     for(i=;i<=n1;i++)
     {
         memset(vis,,sizeof(vis));
         if(hungry(i)==)sum++;
     }
     printf("%d",sum);
     return ;
 }

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。

 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<cstdlib>
 #include<algorithm>
 using namespace std;
 int n,s,t,q[],dis[],G[][];
 int Min(int aaa,int bbb){if(aaa>bbb)return bbb;else return aaa;}
 int Bfs()
 {
     int head,tail,u,v;
     memset(q,,sizeof(q)),head=,tail=,q[tail]=s;
     memset(dis,-,sizeof(dis)),dis[s]=;
     while(head<=tail)
     {
         u=q[++head];
         for(v=;v<=n;v++)
         {
             if(G[u][v]>&&dis[v]<)
             {
                 dis[v]=dis[u]+;
                 q[++tail]=v;
             }
         }
     }
     if(dis[n]<=)return ;
     else return ;
 }
 int Dfs(int u,int minflow)
 {
     int v,ans;
     if(u==n)return minflow;
     for(v=;v<=n;v++)
     {
         if(G[u][v]>&&dis[v]==dis[u]+)
         {
             ans=Dfs(v,Min(minflow,G[u][v]));
             if(ans!=)
             {
                 G[u][v]-=ans;
                 G[v][u]+=ans;
                 return ans;
             }
         }
     }
     return ;
 }
 int Dinic()
 {
     int maxflow=,ans;
     while(Bfs()!=)
     {
         ans=Dfs(s,0x7fffffff);if(ans==)break;maxflow+=ans;
     }
     return maxflow;
 }
 int main()
 {
     int n1,a,b,i;
     scanf("%d %d",&n,&n1);
     while(scanf("%d %d",&a,&b)!=EOF)G[a][b]=;
     n+=;s=n-;t=n;
     for(i=;i<=n1;i++)G[s][i]=;
     for(i=n1+;i<=n-;i++)G[i][t]=;
     printf("%d",Dinic());
     return ;
 }