有源汇有上下界最小流(当前弧优化-->取地址就会超时，赋值秒过)

bdfs后无果

LOJ题面

两份代码不同，求助大佬:

ll dinic(ll u,ll tt,ll flow){
    if(u==tt)return flow;
    ll rst=flow,son;
    for(ll &i=strt[u];i;i=edge[i].nxt){
//  for(ll i=strt[u];i;i=edge[i].nxt){
//      strt[u]=i;
        if(rst==0)break;
        if(deep[edge[i].v]!=deep[u]+1||edge[i].c==0)continue;
        son=dinic(edge[i].v,tt,min(rst,edge[i].c));
        if(son==0)deep[edge[i].v]=0;
        edge[i].c-=son;edge[i^1].c+=son;rst-=son;
    }
    return flow-rst;
}

注释中的代码替换上面的循环后完全无压力通过，上述代码有一个1002ms

评测记录95pts

评测记录AC

全部代码：

#include<cstdio>
#include<cstring>
#define oo 2147483647ll
#define ll long long
using namespace std;
struct node{ll v,c,nxt;}edge[5000006];
ll head[1000006],cnt=1,strt[1000006];
void adge(ll u,ll v,ll c){
    edge[++cnt]=(node){v,c,head[u]};head[u]=cnt;
    edge[++cnt]=(node){u,0,head[v]};head[v]=cnt;
}
ll n,m;
ll S,T;
ll b[4000005],c[4000005];
ll ru[4000005],chu[4000005];
ll u[4000005],v[4000005];
ll Q[1000006],frt,rer;
ll deep[1000004];
bool vis[1000004];
ll min(ll a,ll b){return a<b?a:b;} 
bool bfs(ll ss,ll tt){
    ll u;
    frt=1,rer=0;
    for(ll i=0;i<=n+1;i++)deep[i]=0;
    Q[++rer]=ss;deep[ss]=1;
    while(rer-frt+1){
        u=Q[frt];frt++;
        for(ll i=head[u];i;i=edge[i].nxt){
            if(edge[i].c==0||deep[edge[i].v]!=0)continue;
            deep[edge[i].v]=deep[u]+1;
            Q[++rer]=edge[i].v;
            if(edge[i].v==tt)return true;           
        }
    }
    return false;
}
ll dinic(ll u,ll tt,ll flow){
    if(u==tt)return flow;
    ll rst=flow,son;
    for(ll &i=strt[u];i;i=edge[i].nxt){
//  for(ll i=strt[u];i;i=edge[i].nxt){
//      strt[u]=i;
        if(rst==0)break;
        if(deep[edge[i].v]!=deep[u]+1||edge[i].c==0)continue;
        son=dinic(edge[i].v,tt,min(rst,edge[i].c));
        if(son==0)deep[edge[i].v]=0;
        edge[i].c-=son;edge[i^1].c+=son;rst-=son;
    }
    return flow-rst;
}
int main(){//freopen("a.in","r",stdin);freopen("a.out","w",stdout);
    ll flow;
    ll ans=0;
    ll s,t;
    scanf("%lld%lld%lld%lld",&n,&m,&s,&t);S=0,T=n+1;
    adge(t,s,oo);
    for(ll i=1;i<=m;i++){
        scanf("%lld%lld%lld%lld",&u[i],&v[i],&b[i],&c[i]);
        adge(u[i],v[i],c[i]-b[i]);chu[u[i]]+=b[i];ru[v[i]]+=b[i];
    }
    for(ll i=1;i<=n;i++){
        if(ru[i]-chu[i]>=0)adge(S,i,ru[i]-chu[i]);
        else adge(i,T,chu[i]-ru[i]);
    }
    while(bfs(S,T)){
        for(int i=0;i<=n+1;i++)strt[i]=head[i];
        while(flow=dinic(S,T,oo));
    }
    ans+=edge[3].c;
    edge[2].c=edge[3].c=0;
    for(ll i=head[S];i;i=edge[i].nxt){
        if(edge[i].c!=0)return printf("please go home to sleep\n"),0;
        edge[i].c=edge[i^1].c=0;
    }
    for(ll i=head[T];i;i=edge[i].nxt){
        if(edge[i^1].c!=0)return printf("please go home to sleep\n"),0;
        edge[i].c=edge[i^1].c=0;
    }
    while(bfs(t,s)){
        for(int i=0;i<=n+1;i++)strt[i]=head[i];
        while(flow=dinic(t,s,oo))ans-=flow;
    }
    printf("%lld\n",ans);
    return 0;
}

洛谷上的费用流两份代码都是一样的速度，不太清楚为什么，求调谢谢