模板 有源汇上下界最大流 loj116

时间:2019-11-14
本文章向大家介绍模板 有源汇上下界最大流 loj116,主要包括模板 有源汇上下界最大流 loj116使用实例、应用技巧、基本知识点总结和需要注意事项,具有一定的参考价值,需要的朋友可以参考一下。

加从t到s的流量无穷大,然后在可行流的残留网络上跑最大流。

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int maxn = 210,maxm = 20800,inf = 1000000000;
int cnt = 1,ss,tt,s,t,n,m;
int head[maxn],dis[maxn],d[maxn],nxt[maxm],to[maxm],flow[maxm],low[maxm];
bool inq[maxn]; 
void add(int a,int b,int fl)
{
    nxt[++cnt] = head[a];
    to[cnt] = b;
    head[a] = cnt;
    flow[cnt] = fl;
    nxt[++cnt] = head[b];
    to[cnt] = a;
    head[b] = cnt;
    flow[cnt] = 0;
}
  
bool bfs()
{
    queue <int> que;
    memset(inq,0,sizeof(inq));
    que.push(tt);
    inq[tt] = 1;
    while (!que.empty())
    {
        int cur = que.front();
        que.pop();
        for (int i = head[cur];i;i = nxt[i])
        {
            int v = to[i];
            if (!inq[v] && flow[i ^ 1])
            {
                dis[v] = dis[cur] + 1;
                que.push(v);
                inq[v] = 1;
            }
        }
    }
    return inq[ss];
}
int dfs(int cur,int lmt)
{
    if (cur == tt) return lmt;
    int fl = 0;
    for (int i = head[cur];i && fl < lmt;i = nxt[i])
    {
        if (dis[to[i]] + 1 == dis[cur] && flow[i])
        {
            int tt = dfs(to[i],min(lmt - fl,flow[i]));
            flow[i] -= tt;
            flow[i ^ 1] += tt;
            fl += tt;
        }   
    }
    return fl;
}
int maxflow()
{
    int res = 0;
    while (bfs())
    {
        int fl = 0;
        do
        {
            fl = dfs(ss,inf);
            res += fl;
        } while (fl);
    }
    return res;
}
//cnt = 1
int main()
{
    scanf("%d%d%d%d",&n,&m,&s,&t);
    ss = n + 1;
    tt = n + 2;
    int tx,ty,tl,tu;
    for (int i = 1;i <= m;i++)
    {
        scanf("%d%d%d%d",&tx,&ty,&tl,&tu);
        add(tx,ty,tu - tl);
        d[tx] -= tl;
        d[ty] += tl;
        low[i] = tl;
    }
    add(t,s,inf);
    int sum = 0;
    for (int i = 1;i <= n;i++)
        if (d[i] > 0)
        {
            sum += d[i];
            add(ss,i,d[i]);
        }else if (d[i] < 0)
            add(i,tt,-d[i]); 
    if (maxflow() != sum)
    {
        printf("please go home to sleep\n");
        return 0;
    }else
    {
        ss = s;
        tt = t;
        printf("%d\n",maxflow());
    }
    return 0;
}

原文地址:https://www.cnblogs.com/iat14/p/11858473.html