hdu1158Employment Planning dp 滚动数组

Problem Description

A project manager wants to determine the number of the workers needed in every month. He does know the minimal number of the workers needed in each month. When he hires or fires a worker, there will be some extra cost. Once a worker is hired, he will get the salary even if he is not working. The manager knows the costs of hiring a worker, firing a worker, and the salary of a worker. Then the manager will confront such a problem: how many workers he will hire or fire each month in order to keep the lowest total cost of the project.

 

Input

The input may contain several data sets. Each data set contains three lines. First line contains the months of the project planed to use which is no more than 12. The second line contains the cost of hiring a worker, the amount of the salary, the cost of firing a worker. The third line contains several numbers, which represent the minimal number of the workers needed each month. The input is terminated by line containing a single '0'.

 

Output

The output contains one line. The minimal total cost of the project.

 

Sample Input

   
   
   
   

    
    
    
    3 
4 5 6
10 9 11
0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    199
   
   
   
   

 

Source

Asia 1997, Shanghai (Mainland China) 

Wa了好几次才发现自己真是图样图森破→_→ 开始还自作聪明的把当前人数这个数组改成数字了 怎么不想想真要是照着我这么写 连DP数组也不用写了 ╭(╯^╰)╮

而且 当时只是顾得递推关系中后面的最优解一定是由前面的最优解得来的 而没有注意到这个最优解应该是由几个状态转化而来的 而不是一个状态经过不同操作转化而来的

却说现在还没想到针对于自己之前错误思路的反例 但是理解了正确代码中除了月份循环以外的二重的意义：类似于上一个DP 1024max sumsum plusplus 中滚动数组状态对应取最优的思想

先贴错误代码：

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
long long  hire,salary,fire;
int n;
long long  minn[1000],tmp,dp[10000];///tmp不用设成数组 有数字就行(⊙﹏⊙)b
int main()
{
    while(~scanf("%d",&n)&&n)
    {
        scanf("%lld%lld%lld",&hire,&salary,&fire);
        memset(dp,0,sizeof(dp));
        memset(minn,0,sizeof(minn));
        for(int i=0;i<n;i++) scanf("%lld",&minn[i]);
        dp[0]=(salary+hire)*minn[0];
        tmp=minn[0];
        for(int i=1;i<n;i++)
        {
            if(tmp<minn[i])
            {
                dp[i]=dp[i-1]+(hire)*(minn[i]-tmp)+salary*minn[i];
                tmp=minn[i];
            }
            else if(tmp==minn[i]) dp[i]=dp[i-1]+tmp*salary;
            else
            {
                int tmp1=dp[i-1]+fire*(tmp-minn[i])+salary*minn[i];
                int tmp2=dp[i-1]+salary*tmp;
                if(tmp1<=tmp2) {
                    dp[i]=tmp1;
                    tmp=minn[i];
                }
                else{
                    dp[i]=tmp2;
                }
            }
        }
        printf("%lld\n",dp[n-1]);
    }
    return 0;
}

还是心不静啊啊啊啊啊 都明白啥意思了还能一遍一遍错→_→

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int  hire,salary,fire;
int n;
int  minn[1000],tmp,dp[20][1000],maxn,mmin;///tmp不用设成数组 有数字就行(⊙﹏⊙)b
int main()
{
    while(~scanf("%d",&n)&&n)
    {
        scanf("%d%d%d",&hire,&salary,&fire);
       // memset(dp,0,sizeof(dp));
        //memset(minn,0,sizeof(minn));
        maxn=0;
        for(int i=1;i<=n;i++) {
            scanf("%d",&minn[i]);
            maxn=max(maxn,minn[i]);
        }
        for(int i=minn[1];i<=maxn;i++) dp[1][i]=(salary+hire)*i;
        for(int i=2;i<=n;i++)
        {
            for(int j=minn[i];j<=maxn;j++)//now
            {
                mmin=0x3f3f3f3f;
                for(int k=minn[i-1];k<=maxn;k++)//last
                {
                    if(j>k) tmp=dp[i-1][k]+hire*(j-k)+salary*j;
                    else tmp=dp[i-1][k]+fire*(k-j)+salary*j;
                    if(tmp<mmin) mmin=tmp;
                }
                dp[i][j]=mmin;
            }
        }
        mmin=0x3f3f3f3f;///
        for(int i=minn[n];i<=maxn;i++)
        {
            if(dp[n][i]<mmin) mmin=dp[n][i];
        }
        printf("%d\n",mmin);
    }
    return 0;
}