Storage Keepers - UVa 10163 dp

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本文介绍了一种算法，用于解决公司如何选择最优的守护者来保护一定数量的仓库，并确保在达到最大安全性的同时使支付的工资总额最少。通过二分查找确定最高安全性阈值，再使用动态规划求解最小成本。

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Storage Keepers

Background

Randy Company has N (1<=N<=100) storages. Company wants some men to keep them safe. Now there are M (1<=M<=30) men asking for the job. Company will choose several from them. Randy Company employs men following these rules:

1. Each keeper has a number Pi (1<=Pi<=1000) , which stands for their ability.

2. All storages are the same as each other.

3. A storage can only be lookd after by one keeper. But a keeper can look after several storages. If a keeper’s ability number is Pi, and he looks after K storages, each storage that he looks after has a safe number Uj=Pi div K.(Note: Uj, Pi and K are all integers). The storage which is looked after by nobody will get a number 0.

4. If all the storages is at least given to a man, company will get a safe line L=min Uj

5. Every month Randy Company will give each employed keeper a wage according to his ability number. That means, if a keeper’s ability number is Pi, he will get Pi dollars every month. The total money company will pay the keepers every month is Y dollars.

Now Randy Company gives you a list that contains all information about N,M,P, your task is give company a best choice of the keepers to make the company pay the least money under the condition that the safe line L is the highest.

Input

The input file contains several scenarios. Each of them consists of 2 lines:

The first line consists of two numbers (N and M), the second line consists of M numbers, meaning Pi (I=1..M). There is only one space between two border numbers.

The input file is ended with N=0 and M=0.

Output

For each scenario, print a line containing two numbers L(max) and Y(min). There should be a space between them.

Sample Input

2 1

1 2

10 9

2 5

10 8 6 4 1

5 4

1 1 1 1

0 0

Sample Output

3 7

10 10

8 18

0 0

题意：找到最大的安全值，然后使得花费最小。

思路：先二分找到最大的安全值，然后dp[i][j]表示前i个人保管j个物品时的最小花费，然后递推。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int num[40],n,m,dp[40][110],INF=1000000000;
int solve(int k)
{ if(k==0)
   return 10000;
  int ans=0,i;
  for(i=1;i<=m;i++)
   ans+=num[i]/k;
  return ans;
}
int main()
{ int i,j,k,l,r,mi;
  while(~scanf("%d%d",&n,&m) && n+m)
  { for(i=1;i<=m;i++)
     scanf("%d",&num[i]);
    sort(num+1,num+1+m);
    l=0;r=num[m];
    while(l<r)
    { mi=(l+r+1)/2;
      if(solve(mi)>=n)
       l=mi;
      else
       r=mi-1;
    }
    if(l==0)
    { printf("0 0\n");
      continue;
    }
    for(i=0;i<=m+1;i++)
     for(j=0;j<=n+1;j++)
      dp[i][j]=INF;
    dp[0][0]=0;
    for(i=1;i<=m;i++)
    { k=num[i]/l;
      for(j=n;j>=k;j--)
       dp[i][j]=min(dp[i][j+1],min(dp[i-1][j],dp[i-1][j-k]+num[i]));
      for(j=k;j>=0;j--)
       dp[i][j]=min(dp[i][j+1],min(dp[i-1][j],num[i]));
    }
    printf("%d %d\n",l,dp[m][n]);
  }
}