uva--10163(dp,01背包，双肩包）

10163 Storage Keepers

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.
Sample Input
2 1
7
1 2
10 9
2 5
10 8 6 4 1
5 4
1 1 1 1
0 0

    想到了分成两个子问题，先找出最大的安全值L，再去确定达到L所需要的最小花费，一般既满足什么又得怎么样的时候都得分成两个子问题才行，以前也有个题，说什么长度不小于多少的最大子串和，当时也是分成两个子问题来的，先找最大子串和，然后再DP一次保证长度大于多少，扯远了……但是自己也想到了是背包，可是开始自己胡乱定义状态，明明都想到了背包这种经典模型，为何不直接套用方程呢，自己定义的状态乱七八糟根本不对。

    求这两个子问题的时候都用到了01背包，看网上的名字也很可爱，叫双肩包~

    求解L：

    dp[j]=max( min(dp[j-k],man[i]/k , dp[j]); 选和不选两种情况，k是这个人选择的保护的个数，我现在只能先写成二维的再改成滚动数组，要不不知怎的现在还是一下子没法写出来。

  求解花费：

  dp[j]=min( dp[j-k]+man[i] , dp[j] ) ; k的范围就是man[i]/safe了，这样既满足了第一个条件，又能求出最小的花费。

自己写动归边界问题总是处理不好一直做不对，以后得多加注意。。。啊。这样说估计也没什么用，还得多思考。

代码：

#include<iostream>#include<cstring>#include<cstdio>#include<map>#include<cstring>#include<vector>#include<algorithm>#define INF 0X3f3f3f3f#define mem(a,b) memset(a,b,sizeof(a))using namespace std;typedef long long ll;typedef unsigned long long llu;const int maxd=30+5;const int maxn=100+5;//==========================int man[maxd];int n,m;bool cmp(int a,int b){    return a>b;}int get_L(){    int dp[maxn];    mem(dp,0);    dp[0]=INF;    for(int i=1; i<=m; ++i)        for(int j=n; j>=0; --j)            for(int k=1; k<=j && man[i]>=k; ++k)                dp[j]=max( min(dp[j-k],man[i]/k) , dp[j] );    return dp[n];}int get_w(int safe){    if(safe==0) return 0;    int dp[maxn];    mem(dp,INF);    dp[0]=0;    for(int i=1; i<=m; ++i)        for(int j=n; j>0; --j)        {            int cnt=man[i]/safe;            for(int k=min(j,cnt); k>0; --k)            {                dp[j]=min(dp[j-k]+man[i],dp[j]);            }        }    return dp[n];}int main(){    freopen("1.txt","r",stdin);    while(scanf("%d%d",&n,&m)==2 )    {        if(n==0 && m==0)            break;        for(int i=1; i<=m; ++i)            scanf("%d",&man[i]);        sort(man+1,man+m+1,cmp);        int L=get_L();        printf("%d %d\n",L,get_w(L));    }    return 0;}