# [BZOJ4108][Wf2015]Catering (with source and remittance fee flow)

## Explanation:

There is another condition of this question: every point except point 1 can only pass once
So it's very similar to the last question. The only difficulty of this question is the number of 2-n+1
Since ta has a company, we also need a corresponding point to control the total flow, which just uses 1,n+1+1

Drawing method of original drawing:
Split 2-n+1 points xi,yi
s->1,[0,k],0
1->xi,[0,inf],0
yi->t,[0,inf],0
xi->yi,[1,1],0
For the given cost, if the cost of I - > J is c
yi->xj,[0,inf],c

Then the original drawing is transformed to find the minimum cost and maximum flow

## code:

```#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define INF 1e9
const int N=100000;
int tot=-1,point[N],v[N],remind[N],nxt[N],last[N],c[N],dis[N],mincost,d[N];
bool vis[N];
void addline(int x,int y,int cap,int cc)
{
++tot; nxt[tot]=point[x]; point[x]=tot; v[tot]=y; remind[tot]=cap; c[tot]=cc;
++tot; nxt[tot]=point[y]; point[y]=tot; v[tot]=x; remind[tot]=0; c[tot]=-cc;
}
{
int now=t,ans=INF;
while (now!=s)
{
ans=min(ans,remind[last[now]]);
now=v[last[now]^1];
}
now=t;
while (now!=s)
{
remind[last[now]]-=ans;
remind[last[now]^1]+=ans;
now=v[last[now]^1];
}
return ans;
}
bool spfa(int s,int t)
{
queue<int>q;
memset(dis,0x7f,sizeof(dis));
dis[s]=0;
memset(vis,0,sizeof(vis));
q.push(s);
while (!q.empty())
{
int x=q.front(); q.pop(); vis[x]=0;
for (int i=point[x];i!=-1;i=nxt[i])
if (dis[v[i]]>dis[x]+c[i] && remind[i])
{
dis[v[i]]=dis[x]+c[i];
last[v[i]]=i;
if (!vis[v[i]]) vis[v[i]]=1,q.push(v[i]);
}
}
if (dis[t]>INF) return 0;
mincost+=flow*dis[t];
return 1;
}
int main()
{
int n,k,x;
scanf("%d%d",&n,&k);
int s=1,t=n+n+3,ss=t+1,tt=ss+1;
memset(point,-1,sizeof(point));
for (int i=1;i<=n;i++)
for (int j=1;j<=n-i+1;j++)
for (int i=1;i<=t;i++)
{
}
while (spfa(ss,tt));
printf("%d",mincost);
}```

Posted on Thu, 30 Apr 2020 23:29:33 -0700 by cobaswosa