hdu 1158 Employment Planning

Employment Planning

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4164    Accepted Submission(s): 1743

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)

 

Recommend

Ignatius

正确转移：

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<climits>
#include<queue>
#include<vector>
#include<map>
#include<sstream>
#include<set>
#include<stack>
#include<utility>
//#pragma comment(linker, "/STACK:102400000,102400000")
#define PI 3.1415926535897932384626
#define eps 1e-10
#define sqr(x) ((x)*(x))
#define FOR0(i,n)  for(int i=0 ;i<(n) ;i++)
#define FOR1(i,n)  for(int i=1 ;i<=(n) ;i++)
#define FORD(i,n)  for(int i=(n) ;i>=0 ;i--)
#define  lson   num<<1,le,mid
#define rson    num<<1|1,mid+1,ri
#define MID   int mid=(le+ri)>>1
#define zero(x)((x>0? x:-x)<1e-15)

using namespace std;

const int maxn=1000+20   ;
//const int maxm=    ;
//const int INF=    ;
//typedef long long ll;
const int INF=0x3f3f3f3f;
//ifstream fin("input.txt");
//ofstream fout("output.txt");
//fin.close();
//fout.close();
//freopen("a.in","r",stdin);
//freopen("a.out","w",stdout);

int dp[14][maxn];
int a[14];
int main()
{
//我从来不贴代码
  int n;int hire,salary,fire;
  while(~scanf("%d",&n)&&n)
  {
      scanf("%d%d%d",&hire,&salary,&fire);
      int maxi=0;
      FOR1(i,n)  scanf("%d",&a[i]),maxi=max(maxi,a[i]);
      a[0]=0;
      for(int i=0;i<=maxi;i++)  dp[0][i]=INF;
      dp[0][0]=0;
      maxi=0;
     for(int i=1;i<=n;i++)
     {
         for(int j=a[i];j<=max(a[i],maxi);j++)
         {
             if(  j>a[i-1] )   dp[i][j]=dp[i-1][a[i-1]]+ hire*(j-a[i-1] )+salary*(j);
             else     dp[i][j]=dp[i-1][a[i-1]]+fire*(a[i-1]-j)+salary*(j);

              for(int k=a[i-1]+1;k<= maxi ;k++)
              {

                  if( j>k  )
                  dp[i][j]=min(dp[i][j],dp[i-1][k]+hire*(j-k)+salary*j);
                  else
                  dp[i][j]=min(dp[i][j],dp[i-1][k]+fire*(k-j)+salary*j);
              }

         }
         maxi=max(a[i],maxi);
     }
     int ans=INT_MAX;
     for(int i=a[n];i<=maxi;i++)
        ans=min(ans,dp[n][i]);
     printf("%d\n",ans);
  }





    return 0;
}

错误转移：

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<climits>
#include<queue>
#include<vector>
#include<map>
#include<sstream>
#include<set>
#include<stack>
#include<utility>
//#pragma comment(linker, "/STACK:102400000,102400000")
#define PI 3.1415926535897932384626
#define eps 1e-10
#define sqr(x) ((x)*(x))
#define FOR0(i,n)  for(int i=0 ;i<(n) ;i++)
#define FOR1(i,n)  for(int i=1 ;i<=(n) ;i++)
#define FORD(i,n)  for(int i=(n) ;i>=0 ;i--)
#define  lson   num<<1,le,mid
#define rson    num<<1|1,mid+1,ri
#define MID   int mid=(le+ri)>>1
#define zero(x)((x>0? x:-x)<1e-15)

using namespace std;

const int maxn=1000+20   ;
//const int maxm=    ;
//const int INF=    ;
//typedef long long ll;
const int INF=0x3f3f3f3f;
//ifstream fin("input.txt");
//ofstream fout("output.txt");
//fin.close();
//fout.close();
//freopen("a.in","r",stdin);
//freopen("a.out","w",stdout);

int dp[14][maxn];
int a[14];
int main()
{
//我从来不贴代码
  int n;int hire,salary,fire;
  while(~scanf("%d",&n)&&n)
  {
      scanf("%d%d%d",&hire,&salary,&fire);
      int maxi=0;
      FOR1(i,n)  scanf("%d",&a[i]),maxi=max(maxi,a[i]);
      a[0]=0;
      for(int i=0;i<=maxi;i++)  dp[0][i]=INF;
      dp[0][0]=0;
      maxi=0;
     for(int i=1;i<=n;i++)
     {
         for(int j=a[i];j<=max(a[i],maxi);j++)
         {
//             if(  j>a[i-1] )   dp[i][j]=dp[i-1][a[i-1]]+ hire*(j-a[i-1] )+salary*(j);
//             else     dp[i][j]=dp[i-1][a[i-1]]+fire*(a[i-1]-j)+salary*(j);
                dp[i][j]=INF;
         }
         for(int k=a[i-1];k<= maxi ;k++)
        {
            if(  k<=a[i])
                dp[i][a[i]]=min(dp[i][a[i]],dp[i-1][k]+hire*(a[i]-k)+salary*a[i]);
            else if( salary<fire  )
                dp[i][k]=min(dp[i][k],dp[i-1][k]+salary*k);
            else if(salary >fire)
                dp[i][a[i]]=min( dp[i][a[i]],dp[i-1][k]+fire*(k-a[i])+salary*(a[i])   );
            else if(salary==fire)
            {
                for(int left=a[i];left<=k;left++ )
                    dp[i][left]=min(dp[i][left],dp[i-1][k]+fire*(k-left)+salary*left    );
            }
/*
                  if( j>k  )
                  dp[i][j]=min(dp[i][j],dp[i-1][k]+hire*(j-k)+salary*j);
                  else
                  dp[i][j]=min(dp[i][j],dp[i-1][k]+fire*(k-j)+salary*j);*/
        }
         maxi=max(a[i],maxi);
     }
     int ans=INT_MAX;
     for(int i=a[n];i<=maxi;i++)
        ans=min(ans,dp[n][i]);
     printf("%d\n",ans);
  }





    return 0;
}