bzoj1070 [SCOI2007] Car repair cost flow + split

At the beginning of this question, it is easy to think of the most violent construction method. Since the previous decision affects the waiting time of the people behind, it is enumerated which number to choose.

Since the back is affected by the front, it can also be regarded as the front impact on the back, and the impact is at most n, so enumerate the impact and build the map. Since the arrangement of 4 of 3 vehicles is definitely not as good as 3, there is no aftereffect. Up

code:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int tot=-1,hou[2000007],hf[10][99],che[99],s,t,qj[2000007],xia[2000007],zhong[2000007],c[2000007],v[2000007],cnt,n,m,i,j,k,ans,d[2000007];
bool vis[2000007];
#define inf 9999999
queue<int>q;
void jian(int a,int b,int C,int d)
{
    ++tot;hou[tot]=xia[a],xia[a]=tot;zhong[tot]=b;c[tot]=d,v[tot]=C;    
}
void jia(int a,int b,int c,int d)
{
    jian(a,b,c,d);
    jian(b,a,0,-d);
}
bool bfs()
{
    qj[s]=-1;
    int i;
    for(i=1;i<=cnt;i++)
    d[i]=inf;
    d[s]=0;q.push(s);
    while(!q.empty())
    {
        int st=q.front();
        q.pop();
        vis[st]=0;
        for(i=xia[st];i!=-1;i=hou[i])
        {
            if(v[i]==0)continue;
            int nd=zhong[i];
          if(d[nd]>d[st]+c[i])
          {qj[nd]=i^1;
            d[nd]=d[st]+c[i];
            if(vis[nd]==0)
            {
            vis[nd]=1;
              q.push(nd);   
            }
          }     
        }
    }
    if(d[t]==inf)return 0;
    int o=qj[t];
    while(o!=-1)
    {
        v[o]++;
        v[o^1]--;
        ans+=c[o^1];
        o=qj[zhong[o]];
    }
    return 1;
}
int mcmf()
{
while(bfs());
return ans; 
}
int main()
{
    memset(xia,-1,sizeof(xia));
    scanf("%d%d",&m,&n);
    for(i=1;i<=n;i++)
    for(j=1;j<=m;j++)
    scanf("%d",&hf[j][i]);
    cnt=m;
    for(i=1;i<=m;i++)//枚举人 
    {   
    for(j=1;j<=n;j++)//枚举第几个 
    {
        int o=++cnt;
jia(i,o,1,0);
    for(k=1;k<=n;k++)//枚举是第几辆
    {
        if(che[k]==0)che[k]=++cnt;
        jia(o,che[k],1,hf[i][k]*j); 
    }
    }
    }
    s=++cnt;t=++cnt;
    for(i=1;i<=m;i++)jia(s,i,inf,0);
    for(i=1;i<=n;i++)jia(che[i],t,1,0);
    printf("%.2lf",1.00*mcmf()/n);
}