Loj #116. 有源汇有上下界最大流

题意如文章标题。。。

在有源汇可行流上从S到T再跑一遍最大流，可行流中T->S的流量+这次的最大流流量就是答案。

#include<bits/stdc++.h>
#define ll long long
#define pb push_back
#define maxn 305
using namespace std;
const int inf=1<<30;
struct lines{
	int to,flow,cap;
}l[100005];
vector<int> g[maxn];
int t=-1,n,m,S,T,SS,TT;
int inside[maxn],mx,tot=0;
int cur[maxn],d[maxn];
bool v[maxn],flag=0;

inline void add(int from,int to,int cap){
	l[++t]=(lines){to,0,cap},g[from].pb(t);
	l[++t]=(lines){from,0,0},g[to].pb(t);
}

inline bool BFS(int start){
	memset(v,0,sizeof(v));
	queue<int> q;
	int x; lines e;
	d[start]=0,v[start]=1,q.push(start);
	
	while(!q.empty()){
		x=q.front(),q.pop();
		
		for(int i=g[x].size()-1;i>=0;i--){
			e=l[g[x][i]];
			if(flag&&g[x][i]>mx) continue;
			
			if(e.flow<e.cap&&!v[e.to]){
				v[e.to]=1,d[e.to]=d[x]+1;
				q.push(e.to);
			}
		}
	}
	
	return flag?v[T]:v[TT];
}

int dfs(int x,int a){
	if(!a||(flag&&x==T)||(!flag&&x==TT)) return a;
	int flow=0,f,sz=g[x].size();
	for(int &i=cur[x];i<sz;i++){
		if(flag&&g[x][i]>mx) continue;
		
		lines &e=l[g[x][i]];
		if(d[e.to]==d[x]+1&&(f=dfs(e.to,min(a,e.cap-e.flow)))){
			flow+=f,a-=f;
			e.flow+=f,l[g[x][i]^1].flow-=f;
			if(!a) break;
		}
	}
	
	return flow;
}

inline bool can(){
	int an=0;
	while(BFS(SS)){
		memset(cur,0,sizeof(cur));
		an+=dfs(SS,inf);
	}
	
	return an==tot;
}

inline int calc(){
	int an=0;
	while(BFS(S)){
		memset(cur,0,sizeof(cur));
		an+=dfs(S,inf);
	}
	
	return an;
}

int main(){
	scanf("%d%d%d%d",&n,&m,&S,&T);
	int uu,vv,bt,cl;
	SS=0,TT=n+1;
	for(int i=1;i<=m;i++){
		scanf("%d%d%d%d",&uu,&vv,&bt,&cl);
		add(uu,vv,cl-bt);
		inside[vv]+=bt;
		inside[uu]-=bt;
	}
	mx=t;
	for(int i=1;i<=n;i++){
		if(inside[i]>0) add(SS,i,inside[i]),tot+=inside[i];
		else add(i,TT,-inside[i]);
	}
	add(T,S,inf);
	
	if(!can()){
		puts("please go home to sleep");
		return 0;
	}
	
	flag=1;
	
	printf("%d
",l[t-1].flow+calc());
	return 0;
}