题目链接：https://loj.ac/problem/117

解法：

（1）增加超级源点st和超级汇点sd，对于有上下界的边（i，j）流量（L,R）变为R-L，然后i与sd连接容量是L，st与j连接容量是L；网络中规定不能有流量流入st，也不能有流量流入sd；

（2）做一次最大流Dinic；

（3）在汇点t到s连一条容量是inf的边；

（4）在做一次最大流Dinic

（5）当且仅当附加弧都满流是有可行流，最后的最小流是flow[sd->st]^1],st到sd的最大流就是sd到st的最小流；

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL inf = 1e10;
const LL maxn = 70005;
const LL maxm = 125005;
struct G
{
    LL v, cap, next, num;
    G() {}
    G(LL v, LL cap, LL next, LL num) : v(v), cap(cap), next(next), num(num) {}
} E[maxm*10];
LL p[maxn], T;
LL d[maxn], temp_p[maxn], qw[maxn]; //d顶点到源点的距离标号,temp_p当前狐优化，qw队列
LL n,m,s,t;
void init()
{
    memset(p, -1, sizeof(p));
    T = 0;
}
void add(LL u, LL v, LL cap, LL num)
{
    E[T] = G(v, cap, p[u], num);
    p[u] = T++;
    E[T] = G(u, 0, p[v], num);
    p[v] = T++;
}
bool bfs(LL st, LL en, LL n)
{
    LL i, u, v, head, tail;
    for(i = 0; i <= n; i++) d[i] = -1;
    head = tail = 0;
    d[st] = 0;
    qw[tail] = st;
    while(head <= tail)
    {
        u = qw[head++];
        for(i = p[u]; i + 1; i = E[i].next)
        {
            v = E[i].v;
            if(d[v] == -1 && E[i].cap > 0)
            {
                d[v] = d[u] + 1;
                qw[++tail] = v;
            }
        }
    }
    return (d[en] != -1);
}
LL dfs(LL u, LL en, LL f)
{
    if(u == en || f == 0) return f;
    LL flow = 0, temp;
    for(; temp_p[u] + 1; temp_p[u] = E[temp_p[u]].next)
    {
        G& e = E[temp_p[u]];
        if(d[u] + 1 == d[e.v])
        {
            temp = dfs(e.v, en, min(f, e.cap));
            if(temp > 0)
            {
                e.cap -= temp;
                E[temp_p[u] ^ 1].cap += temp;
                flow += temp;
                f -= temp;
                if(f == 0)  break;
            }
        }
    }
    return flow;
}
LL dinic(LL st, LL en, LL n)
{
    LL i, ans = 0;
    while(bfs(st, en, n))
    {
        for(i = 0; i <= n; i++) temp_p[i] = p[i];
        ans += dfs(st, en, inf);
    }
    return ans;
}
LL du[maxn],sum=0, ans=0;
int main()
{
    scanf("%lld%lld%lld%lld",&n,&m,&s,&t);
    init();
    memset(du,0,sizeof(du));
    for(LL i=1; i<=m; i++){
        LL u,v,l,r;
        scanf("%lld %lld %lld %lld", &u,&v,&l,&r);
        add(u,v,r-l, i);
        du[u]-=l;
        du[v]+=l;
    }
    LL ss = 0, tt = n+1;
    for(LL i=1; i<=n; i++){
        if(du[i]>0) sum+=du[i],add(ss,i,du[i], 0);
        if(du[i]<0) add(i,tt,-du[i], 0);
    }
    ans += dinic(ss, tt, n+2);
    add(t, s, inf, 0);
    ans += dinic(ss, tt, n+2);
    if(sum == ans){
        printf("%lld\n", E[T-1].cap);
    }
    else{
        puts("please go home to sleep");
    }
    return 0;
}