题目链接:点击查看
题目大意:给出一个 n 个点和 m 条边的有向图,每条边都有一个流量限制 [ lower , upper ],给定源点 s 和汇点 t ,求出源点到汇点的最小流
题目分析:参考我的上一篇博客:https://blog.csdn.net/qq_45458915/article/details/108339354
坑点就是卡常+卡当前弧优化,第八个点会TLE,如果有当前弧优化可以删除试一下,如果链式前向星是用结构体封装的可以解封装试一下
代码:
#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
#include<bitset>
#include<unordered_map>
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
const int inf=0x3f3f3f3f;
const int N=50100;
const int M=125100;
int du[N];
int head[N],cnt,edge[(N+M)<<1],ver[(N+M)<<1],nt[(N+M)<<1];
void addedge(int u,int v,int w)
{
ver[cnt]=v;
edge[cnt]=w;
nt[cnt]=head[u];
head[u]=cnt++;
ver[cnt]=u;
edge[cnt]=0;//反向边边权设置为0
nt[cnt]=head[v];
head[v]=cnt++;
}
int d[N];//深度
bool bfs(int s,int t)//寻找增广路
{
memset(d,0,sizeof(d));
queue<int>q;
q.push(s);
d[s]=1;
while(!q.empty())
{
int u=q.front();
q.pop();
for(int i=head[u];i!=-1;i=nt[i])
{
if(d[ver[i]])
continue;
if(!edge[i])
continue;
d[ver[i]]=d[u]+1;
q.push(ver[i]);
if(ver[i]==t)
return true;
}
}
return false;
}
int dinic(int x,int t,int flow)//更新答案
{
if(x==t)
return flow;
int rest=flow,i;
for(i=head[x];i!=-1&&rest;i=nt[i])
{
if(edge[i]&&d[ver[i]]==d[x]+1)
{
int k=dinic(ver[i],t,min(rest,edge[i]));
if(!k)
d[ver[i]]=0;
edge[i]-=k;
edge[i^1]+=k;
rest-=k;
}
}
return flow-rest;
}
void init()
{
memset(head,-1,sizeof(head));
cnt=0;
}
int solve(int st,int ed)
{
int ans=0,flow;
while(bfs(st,ed))
while(flow=dinic(st,ed,inf))
ans+=flow;
return ans;
}
int main()
{
#ifndef ONLINE_JUDGE
// freopen("data.in.txt","r",stdin);
// freopen("data.out.txt","w",stdout);
#endif
// ios::sync_with_stdio(false);
init();
int n,m,s,t;
scanf("%d%d%d%d",&n,&m,&s,&t);
while(m--)
{
int u,v,lower,upper;
scanf("%d%d%d%d",&u,&v,&lower,&upper);
addedge(u,v,upper-lower);
du[u]-=lower,du[v]+=lower;
}
int st=N-1,ed=st-1,sum=0;
for(int i=1;i<=n;i++)
{
if(du[i]>0)
{
addedge(st,i,du[i]);
sum+=du[i];
}
else
addedge(i,ed,-du[i]);
}
addedge(t,s,inf);
if(solve(st,ed)!=sum)
{
puts("please go home to sleep");
return 0;
}
int ans=edge[cnt-1];
edge[cnt-1]=edge[cnt-2]=0;
printf("%d\n",ans-solve(t,s));
return 0;
}