bzoj1070[SCOI2007]修车

传送门

最小费用最大流，又是一道思维题，我又没想出建图，问题想对了，就是不会解决，建图大概就是对于每个工人每个时间段建个点，只要想到将(n)个工人分成((n*m))个点，剩下的就简单了，zkw费用流在这题表现不佳

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
inline void read(int &x) {
    char ch; bool ok;
    for(ok=0,ch=getchar(); !isdigit(ch); ch=getchar()) if(ch=='-') ok=1;
    for(x=0; isdigit(ch); x=x*10+ch-'0',ch=getchar()); if(ok) x=-x;
}
#define min(a,b) (a<b?a:b)
#define rg register
int n,m,s,t,sla[100001],pre[100001],dis[100001],nxt[100001],h[50001],v[100001],w[100001],cnt=1,inf=1e9,ans;bool vis[100001];
void add(int x,int y,int z,int u)
{
    pre[++cnt]=y,nxt[cnt]=h[x],h[x]=cnt,v[cnt]=z,w[cnt]=u,
    pre[++cnt]=x,nxt[cnt]=h[y],h[y]=cnt,v[cnt]=0,w[cnt]=-u;
}
int dfs(int x,int flow)
{
    if(x==t){ans+=dis[t]*flow;return flow;}
    vis[x]=1;int f=flow;
    for(rg int i=h[x];i;i=nxt[i])
        if(!vis[pre[i]]&&v[i])
        {
            if(!(dis[x]+w[i]-dis[pre[i]]))
            {
                int y=dfs(pre[i],min(v[i],f));
                f-=y,v[i]-=y,v[i^1]+=y;
                if(!f)return flow;
            }
            else sla[pre[i]]=min(sla[pre[i]],dis[x]+w[i]-dis[pre[i]]);
        }
    return flow-f;
}
int aug()
{
    int mn=inf;
    for(rg int i=s;i<=t;i++)if(!vis[i])mn=min(mn,sla[i]),sla[i]=inf;
    if(mn==inf)return 1;
    for(rg int i=s;i<=t;i++)if(!vis[i])dis[i]+=mn;
    return 0;
}
int main()
{
    read(n),read(m),s=0,t=(n+2)*m+1;
    for(rg int i=1,x;i<=m;i++)
        for(rg int j=1;j<=n;j++)
        {
            read(x);
            for(rg int k=1;k<=m;k++)add(i,j*m+k,1,k*x);
        }
    for(rg int i=1;i<=m;i++)add(s,i,1,0);
    for(rg int i=1;i<=m;i++)for(rg int j=1;j<=n;j++)add(j*m+i,t,1,0);
    memset(sla,63,sizeof sla);
    while(1)
    {
        while(1){memset(vis,0,sizeof vis);if(!dfs(s,inf))break;}
        if(aug())break;
    }
    printf("%.2lf
",1.0*ans/m);
}