【洛谷2053】 [SCOI2007]修车（费用流）

传送门

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
",(double)(McMf()*1./n));
	return 0;
}