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洛谷

Solution

考虑把每一个修车工人拆成\(n\)个点，那么考虑令\(id(i,j)\)为第\(i\)个工人倒数第\(j\)次修车。

然后就可以直接跑费用流了！！！

代码实现

/*

  mail: mleautomaton@foxmail.com

  author: MLEAutoMaton

  This Code is made by MLEAutoMaton

*/

#include<stdio.h>

#include<stdlib.h>

#include<string.h>

#include<math.h>

#include<algorithm>

#include<queue>

#include<set>

#include<map>

#include<iostream>

using namespace std;

#define ll long long

#define re register

#define file(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout)

inline int gi()

{

	int f=1,sum=0;char ch=getchar();

	while(ch>'9' || ch<'0'){if(ch=='-')f=-1;ch=getchar();}

	while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}

	return f*sum;

}

int m,n,s,t,T[110][110];

queue<int>Q;

const int N=500010,M=2000010,Inf=1e9+10;

struct node

{

	int to,nxt,w,c;

}e[M<<1];

int front[N],cnt,dis[N],vis[N],fa[N],from[N];

void Add(int u,int v,int flow,int cost)

{

	e[cnt]=(node){v,front[u],flow,cost};front[u]=cnt++;

	e[cnt]=(node){u,front[v],0,-cost};front[v]=cnt++;

}

bool SPFA()

{

	memset(dis,63,sizeof(dis));

	Q.push(s);dis[s]=0;

	while(!Q.empty())

	{

		int u=Q.front();Q.pop();vis[u]=0;

		for(int i=front[u];i!=-1;i=e[i].nxt)

		{

			int v=e[i].to;

			if(e[i].w && dis[v]>dis[u]+e[i].c)

			{

				dis[v]=dis[u]+e[i].c;fa[v]=u,from[v]=i;

				if(!vis[v])Q.push(v),vis[v]=1;

			}

		}

	}

	return dis[t]!=dis[t+1];

}

int McMf()

{

	int cost=0;

	while(SPFA())

	{

		int di=Inf;

		for(int i=t;i!=s;i=fa[i])di=min(di,e[from[i]].w);

		cost+=di*dis[t];

		for(int i=t;i!=s;i=fa[i])

			e[from[i]].w-=di,e[from[i]^1].w+=di;

	}

	return cost;

}

int main()

{

#ifndef ONLINE_JUDGE

	freopen("in.in","r",stdin);

#endif

	m=gi();n=gi();

	memset(front,-1,sizeof(front));

	for(int i=1;i<=n;i++)

		for(int j=1;j<=m;j++)

			T[i][j]=gi();

	for(int i=1;i<=n;i++)

		Add(s,i,1,0);

	t=n*m+n+1;

	for(int i=1;i<=n;i++)

		for(int j=1;j<=m;j++)

			for(int k=1;k<=n;k++)

				Add(i,(j-1)*n+k+n,1,k*T[i][j]);

	for(int j=1;j<=m;j++)

		for(int i=1;i<=n;i++)

			Add((j-1)*n+i+n,t,1,0);

	printf("%.2lf\n",(double)(McMf()*1./n));

	return 0;

}