https://www.luogu.org/problemnew/show/P3254
思路：这道题很水，s向每个单位连弧，容量为单位的人数，每个餐桌向t连弧，容量为餐桌容量，每个单位向每个餐桌连容量为1的弧，跑个最大流就好了。

#include <bits/stdc++.h>
using namespace std;
const int maxn=1005;
const int INF=0x3f3f3f3f;

struct Edge{
	int from,to,cap,flow;
};

struct Dinic{
	int n1,n2,m,s,t,sum;
	vector<Edge> edges;
	vector<int> G[maxn];
	bool vis[maxn];
	int d[maxn];
	int cur[maxn];

	void init()
	{
		int x;
		cin>>n1>>n2;
		s=0;t=n1+n2+1;
		for(int i=1;i<=n1;i++)
		{
			scanf("%d",&x);
			sum+=x;
			AddEdge(s,i,x);
			for(int j=n1+1;j<=n1+n2;j++)AddEdge(i,j,1);
		}
		for(int i=n1+1;i<=n1+n2;i++)
		{
			scanf("%d",&x);
			AddEdge(i,t,x);
		}
	}

	void AddEdge(int f,int t,int c)
	{
		edges.push_back((Edge){f,t,c,0});
		edges.push_back((Edge){t,f,0,0});
		m=edges.size();
		G[f].push_back(m-2);
		G[t].push_back(m-1);
	}

	bool bfs()
	{	
		memset(vis,0,sizeof(vis));
		queue<int> Q;
		Q.push(s);
		d[s]=0;
		vis[s]=1;
		while(!Q.empty())
		{
			int x=Q.front();Q.pop();
			for(int i=0;i<G[x].size();i++)
			{
				Edge& e=edges[G[x][i]];
				if(!vis[e.to] && e.cap>e.flow)
				{
					vis[e.to]=1;
					d[e.to]=d[x]+1;
					Q.push(e.to);
				}
			}
		}
		return vis[t];
	}

	int dfs(int x,int a)
	{
		if(x==t || a==0)return a;
		int flow=0,f;
		for(int& i=cur[x];i<G[x].size();i++)
		{
			Edge& e=edges[G[x][i]];
			if(d[x]+1==d[e.to] && (f=dfs(e.to,min(a,e.cap-e.flow)))>0)
			{
				e.flow+=f;
				edges[G[x][i]^1].flow-=f;
				flow+=f;
				a-=f;
				if(!a)break;
			}
		}
		return flow;
	}

	int MaxFlow()
	{
		int flow=0;
		while(bfs())
		{
			memset(cur,0,sizeof(cur));
			flow+=dfs(s,INF);
		}
		return flow;
	}

	void print(int flow)
	{
		printf("%d\n",flow==sum);
		if(flow!=sum)return;
		int now=1;
		for(int i=0;i<m;i+=2)
		{
			int from=edges[i].from,to=edges[i].to;
			if(from==s||from==t||to==s||to==t)continue;
			if(from!=now)now++,putchar('\n');
			if(edges[i].flow==1)printf("%d ",to-n1);
		}
	}	
}ans;

int main()
{
	//freopen("input.in","r",stdin);
	ans.init();
	int flow=ans.MaxFlow();
	ans.print(flow);
	return 0;
}