LOJ#117. 有源汇有上下界最小流

https://loj.ac/problem/117
模板题。原理分析见https://blog.youkuaiyun.com/axuan_k/article/details/47282421

```#include <bits/stdc++.h>
using namespace std;
const int maxn=60000;
const int INF=0x3f3f3f3f;

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

struct Dinic{
	int n,e,m,s,t,S,T;
	vector<Edge> edges;
	vector<int> G[maxn];
	bool vis[maxn];
	int d[maxn];
	int cur[maxn];
	int sum;

	void init()
	{
		cin>>n>>e>>S>>T;
		s=0,t=n+1;
		int degree[maxn]={0};
		int u,v,b,c;
		for(int i=1;i<=e;i++)
		{
			cin>>u>>v>>b>>c;
			AddEdge(u,v,c-b);
			degree[v]+=b;
			degree[u]-=b;
		}
		for(int i=1;i<=n;i++)
		{
			if(degree[i]>0)AddEdge(s,i,degree[i]),sum+=degree[i];
			else AddEdge(i,t,-degree[i]);
		}
		AddEdge(T,S,INF);
	}

	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 ans)
	{		
		if(ans!=sum){cout<<"please go home to sleep"<<endl;return;}
		ans=edges[m-2].flow;
		G[S].pop_back();G[T].pop_back();
		s=T,t=S;
		int flow=MaxFlow();
		cout<<ans-flow<<endl;
	}
}ans;

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