有源汇有上下界最小流 DInic + 各种优化 模板

例题：loj117 ： https://loj.ac/problem/117

//其实就是判断可行流后倒着求一遍最大流 




#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int maxn = 200010, INF = 0x7fffffff;
int d[maxn], head[maxn], in[maxn], cur[maxn];
int n, m, s, t;
int cnt = 0;

struct node{
    int u, v, c, next, bz;
}Node[2*maxn];

inline void add_(int u, int v, int c, int bz)
{
    Node[cnt].u = u;
    Node[cnt].v = v;
    Node[cnt].c = c;
    Node[cnt].next = head[u];
    Node[cnt].bz = bz;
    head[u] = cnt++;

}
inline void add(int u, int v, int c, int bz)
{
    add_(u,v,c,bz);
    add_(v,u,0,bz);
}

inline bool bfs()  //此处为bool类型  其它类型会TL
{
    queue<int> Q;
    mem(d,0);
    Q.push(s);
    d[s] = 1;
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        for(int i=head[u]; i!=-1; i=Node[i].next)
        {
            node e = Node[i];
            if(!d[e.v] && e.c > 0)
            {
                d[e.v] = d[e.u] + 1;
                Q.push(e.v);
                if(e.v == t) return 1;
            }
        }
    }
    return d[t] != 0;
}

inline int dfs(int u, int cap)
{
    int ret = 0, V;
    if( u == t || cap == 0)
        return cap;
    for(int &i=cur[u]; i!=-1; i=Node[i].next)     //千万不要忘记加引用符号
    {
        node e = Node[i];
        if(d[e.v] == d[e.u] + 1 && e.c > 0)
        {
            int V = dfs(e.v, min(cap, e.c));

            Node[i].c -= V;
            Node[i^1].c += V;
            ret += V;
            cap -= V;
            if(cap == 0) break;

        }
    }
    if(cap > 0) d[u] = -1;
    return ret;
}

inline int Dinic(int u)
{
    int ans = 0;
    while(bfs())
    {
        memcpy(cur,head,sizeof(head));
        ans += dfs(u, INF);
    }
    return ans;
}

int main()
{
    mem(head,-1);
    int s_, t_;
    scanf("%d%d%d%d",&n,&m,&s_,&t_);
    s = 0; t = n+1;
    int sum = 0;
    for(int i=0; i<m; i++)
    {
        int u, v, b, d;
        scanf("%d%d%d%d",&u,&v,&b,&d);
        add(u, v, d-b,1);
        in[v] += b;
        in[u] -= b;
    }
    for(int i=1; i<=n; i++)
    {
        if(in[i] > 0)
        {
            sum += in[i];
            add(s,i,in[i],0);
        }
        else
        {
            add(i,t,-in[i],0);
        }
    }
    add(t_,s_,INF,0);
    if(sum != Dinic(s))
    {
        printf("please go home to sleep
");
    }
    else
    {
        sum = Node[head[t_]^1].c;
        for(int i=0; i<cnt; i++)
        {
            if(!Node[i].bz) Node[i].v = 0;
        }
        head[s] = head[t] = -1;
        s = t_;
        t = s_;
        printf("%d
",sum - Dinic(s));   //sum 减去 反向最大流 = 正向最小流

    }
    return 0;
}