洛谷P2763 试题库问题 最大流

Code:

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<cstring>
using namespace std;
const int maxn=3000;
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) {}
};
vector<Edge>edges;
vector<int>G[maxn];
struct Dicnic{
   int d[maxn],vis[maxn],cur[maxn];
   queue<int>Q;
   void addedge(int u,int v,int c){
   	edges.pb(Edge(u,v,c));               //正向弧
   	edges.pb(Edge(v,u,0));               //反向弧
   	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 k,n;scanf("%d%d",&k,&n);
	s=0,t=k+n+1;
	int sum=0;
	for(int i=1;i<=n;++i)op.addedge(s,i,1);
	for(int i=1;i<=k;++i)
	{
		int a;scanf("%d",&a);
		sum+=a;
		op.addedge(n+i,t,a);
	}
	for(int i=1;i<=n;++i)
	{
		int b;scanf("%d",&b);
		while(b--)
		{
			int a;scanf("%d",&a);
			op.addedge(i,a+n,1);
		}
	}
     int F=op.maxflow();
     if(F!=sum){printf("No Solution!
");return 0;}
     for(int u=n+1;u<=n+k;++u)
     {
     	    printf("%d: ",u-n);
     	    int sz=G[u].size();
     	    for(int i=0;i<sz;++i)
     	    {
     	    	   int e=G[u][i];
     	    	   if(e%2==1)
     	    	   {
     	    	   	   if(edges[e].cap==1)printf("%d ",edges[e].to);
     	    	   }
     	    }
     	    printf("
");
     }
     return 0;
}

  

原文地址:https://www.cnblogs.com/guangheli/p/10365891.html