UVA 10163 Storage Keepers 两次dp

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 ÷ 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 = minUj
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.

uva的题目总是屎一样的粘贴，嘛，就当是索引了。

题意就是说给你几个m保安 看守n个仓库，每个保安有pi的能力，看守的仓库数量均分能力值，一个仓库不能被多次看守，问当使仓库看守度最小的那个最大的情况下，该值和使用的能力值之和最小的值。

蛮绕的 理解起来倒是不麻烦

首先就是dp解决看守度最小的仓库值最大的那个值，转移方程也很好写出来

接下来就是dp寻找最少的花费了，那就再次dp，寻找满足条件的最小花费值，注意当0的时候要特判，原因不解释了

代码：

#include<cstring>#include<cstdio>#include<algorithm>#include<iostream>#include<queue>using namespace std;const int inf=0x3f3f3f3f;int a[150];int dp[55][105];int g[105][105];int m,n;int main(){while(scanf("%d %d",&n,&m)!=EOF&&(n+m)){memset(dp,0,sizeof(dp));for(int i=1;i<=m;i++)scanf("%d",a+i);for(int i=1;i<=m;i++){dp[i-1][0]=inf;for(int j=1;j<=n;j++){dp[i][j]=dp[i-1][j]; // k=0for(int k=1;k<=j;k++)dp[i][j]=max(dp[i][j],min(dp[i-1][j-k],a[i]/k));}}printf("%d ",dp[m][n]);int sav=dp[m][n];        memset(g,inf,sizeof(g));        if(sav==0) printf("0\n");        else{        for(int i=1;i<=m;i++){g[i-1][0]=0;for(int j=1;j<=n;j++){g[i][j]=g[i-1][j];for(int k=1;k<=j;k++){if(a[i]/k>=sav)g[i][j]=min(g[i][j],g[i-1][j-k]+a[i]);}}}printf("%d\n",g[m][n]);}}return 0;}