Uva--10163（动规，递推）

2014-08-25 10:02:20

 Problem C.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

7

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

思路：这题其实也可以归纳为变相的01背包，对于一个应聘的人选或不选，dp[i][j][k]表示前 i 个人管理 k 个仓库，使得safe线达到 j 的最小薪水和。

　　状态转移方程：dp[i][j][k + m] = min(dp[i - 1][j][k] + p[i] , dp[i - 1][j][k + m])   (k = p[i] / j，表示能加入第 i 个人达到 j 的safe线能多管多少个仓库)

/*************************************************************************
    > File Name: k.cpp
    > Author: Nature
    > Mail: 564374850@qq.com 
    > Created Time: Fri 22 Aug 2014 09:23:15 PM CST
************************************************************************/

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;
const int INF = 1 << 30;

int N,M;
int p[35];
int sum[35];
int dp[35][1005][105];

int main(){
    while(scanf("%d%d",&N,&M) == 2){
        if(!N && !M) break;
        for(int i = 1; i <= M; ++i){
            scanf("%d",&p[i]);
            sum[i] = (i == 1 ? p[i] : sum[i - 1] + p[i]);
        }
        int top = sum[min(N,M)] / N;
        if(top == 0){
            printf("0 0\n");
            continue;
        }
        for(int i = 0; i <= M; ++i){
            for(int j = 0; j <= top; ++j){
                for(int n = 0; n <= N; ++n){
                    dp[i][j][n] = INF;
                }
                dp[i][j][0] = 0;
            }
        }
        for(int i = 1; i <= M; ++i)
            for(int j = 1; j <= top; ++j)
                for(int n = 0; n + p[i] / j <= N; ++n)
                    dp[i][j][n + p[i] / j] = min(dp[i - 1][j][n] + p[i],dp[i - 1][j][n + p[i] / j]);
                    
        for(int j = top; j >= 1; --j){
            if(dp[M][j][N] != INF){
                printf("%d %d\n",j,dp[M][j][N]);
                break;
            }
        }
    }
    return 0;
}