1070: [SCOI2007]修车

/*
一开始以为是个贪心 发现自己太naive了

将每个技术工人拆成n个点，一共拆n*m个，第i个表示倒数第i次修车。 
让每辆车向拆出来的点连边，费用为tmp[i][j]*k，i是技工，j是车，k是拆出来的第几个点，
这样设置费用的原因是j是i倒数第k个修的，那么i修的后k个车都要等倒数第k个车。
然后跑最小费用最大流就可以了
*/

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
using namespace std;

inline int read()
{
    char c=getchar();int num=0;
    for(;!isdigit(c);c=getchar());
    for(;isdigit(c);c=getchar())
        num=num*10+c-'0';
    return num;
}

const int N=65;
const int M=2e5+5;

int m,n,S,T;
int tim[N][N];
long long ans;

int head[M],num_edge;
struct Edge
{
    int v,nxt,flow,cost;
}edge[M];

inline void add_edge(int u,int v,int flow,int cost)
{
    edge[++num_edge].v=v;
    edge[num_edge].flow=flow;
    edge[num_edge].cost=cost;
    edge[num_edge].nxt=head[u];
    head[u]=num_edge;
    edge[++num_edge].v=u;
    edge[num_edge].flow=0;
    edge[num_edge].cost=-cost;
    edge[num_edge].nxt=head[v];
    head[v]=num_edge;
}

int dis[M];
queue<int> que;
bool inque[M];
bool spfa()
{
    memset(dis,0x3f,sizeof(dis));
    que.push(S),dis[S]=0;
    int now;
    while(!que.empty())
    {
        now=que.front(),que.pop();
        for(int i=head[now],v;i;i=edge[i].nxt)
        {
            if(edge[i].flow==0)
                continue;
            v=edge[i].v;
            if(dis[v]>dis[now]+edge[i].cost)
            {
                dis[v]=dis[now]+edge[i].cost;
                if(!inque[v])
                {
                    que.push(v);
                    inque[v]=1;
                }
            }
        }
        inque[now]=0;
    }
    return dis[T]!=0x3f3f3f3f;
}

int vis[M],visf;
int dfs(int u,int flow)
{
    if(u==T||!flow)
        return flow;
    int outflow=0;
    vis[u]=visf;
    for(int i=head[u],v,tmp;i;i=edge[i].nxt)
    {
        if(!edge[i].flow)
            continue;
        v=edge[i].v;
        if(vis[v]!=visf&&dis[v]==dis[u]+edge[i].cost)
        {
            tmp=dfs(v,min(flow,edge[i].flow));
            if(!tmp)
                continue;
            ans+=1ll*tmp*edge[i].cost;
            edge[i].flow-=tmp;
            edge[i^1].flow+=tmp;
            flow-=tmp;
            outflow+=tmp;
            if(!flow)
            {
                vis[u]=0;
                return outflow;
            }
        }
    }
    vis[u]=0;
    dis[u]=0x7fffffff;
    return outflow;
}

int main()
{
    num_edge=1;
    m=read(),n=read();
    T=n+n*m+1;
    for(int i=1;i<=n;++i)
        for(int j=1;j<=m;++j)
            tim[j][i]=read();
    for(int i=1;i<=n;++i)
        add_edge(S,i,1,0);
    for(int i=1,p;i<=m;++i)
    {
        for(int j=1;j<=n;++j)
        {
            p=i*n+j;
            add_edge(p,T,1,0);
            for(int c=1;c<=n;++c)
                add_edge(c,p,1,tim[i][c]*j);
        }
    }
    while(spfa())
    {
        ++visf;
        dfs(S,0x7fffffff);
    }
    printf("%.2lf",1.0*ans/n);
    return 0;
}