洛谷P1231 教辅的组成 最大流

裸题…
Code：

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<vector>

#include<cstring>

#include<queue>

using namespace std;

const int maxn=50004;

const int INF=10000000;

# define  pb push_back

int s,t;

struct Edge{

  int from,to,cap;

  Edge(int u,int v,int c):from(u),to(v),cap(c) {}

};

struct Dicnic{

   vector<Edge>edges;

   vector<int>G[maxn];

   int d[maxn],vis[maxn],cur[maxn];

   queue<int>Q;

   void addedge(int u,int v,int c){

    edges.pb(Edge(u,v,c));

    edges.pb(Edge(v,u,0));

    int m=edges.size();

    G[u].pb(m-2);

    G[v].pb(m-1);

   }

   int BFS()

   {

    memset(vis,0,sizeof(vis));

    d[s]=0,vis[s]=1;Q.push(s);

    while(!Q.empty()){

      int u=Q.front();Q.pop();

      int sz=G[u].size();

      for(int i=0;i<sz;++i){

        Edge e=edges[G[u][i]];

        if(!vis[e.to]&&e.cap>0){

          d[e.to]=d[u]+1,vis[e.to]=1;

          Q.push(e.to);

        }

      }

    }

    return vis[t];

   }

   int dfs(int x,int a){

       if(x==t)return a;

       int sz=G[x].size();

       int f,flow=0;

       for(int i=cur[x];i<sz;++i){

        Edge e=edges[G[x][i]];

        cur[x]=i;

        if(d[e.to]==d[x]+1&&e.cap>0){

          f=dfs(e.to,min(a,e.cap));

          if(f)

          {

                int u=G[x][i];

                a-=f;

                edges[u].cap-=f;

                edges[u^1].cap+=f;

                flow+=f;

                if(a==0)break;

          }

        }

       }

       return flow;

   }

   int maxflow(){

    int ans=0;

    while(BFS()){

      memset(cur,0,sizeof(cur));

      ans+=dfs(s,INF);

    }

    return ans;

   }

}op;

int main()

{

	int n1,n2,n3,m;

	scanf("%d%d%d",&n1,&n2,&n3);

	scanf("%d",&m);

	s=0,t=n2+n1*2+n3+3;

	while(m--){

		int a,b;scanf("%d%d",&a,&b);

          op.addedge(b,n2+a,1);

	}                                              //练习册和书

	for(int i=1;i<=n2;++i)op.addedge(s,i,1);              //源点到练习册

	for(int i=n2+1;i<=n2+n1;++i)op.addedge(i,i+n1,1);      //书的容量限制

	scanf("%d",&m);

     while(m--){

     	     int a,b;

     	     scanf("%d%d",&a,&b);

     	     op.addedge(n2+a+n1,n2+n1+n1+b,1);           //书到答案

     }

     for(int i=n2+2*n1+1;i<=n2+2*n1+n3;++i)

     	     op.addedge(i,t,1);                           //答案到汇点

     	int ans=op.maxflow();

     	printf("%d",ans);

     	return 0;

}