Employment planning[hdu1158]

Employment Planning

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 5292 Accepted Submission (s): 2262

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'll be is some extra cost. Once A worker is hired, he'll 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 would confront such a problem:how many workers he'll hire or fire each month in order to keep the Lowes T 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 are 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

#include <stdio.h>#include<string.h>Const intinf=99999999;intdp[ the][10010];intpeople[ the];intMain () {//freopen ("Input.txt", "R", stdin);    intN; intHire,salary,fire;  while(SCANF ("%d", &n) &&N) {scanf ("%d%d%d",&hire,&salary,&Fire ); intmax_people=0; inti,j,k;  for(i=1; i<=n;i++) {scanf ("%d",&People[i]); if(max_people<People[i]) max_people=People[i]; }         for(i=people[1];i<=max_people;i++)//Initialize first monthdp[1][i]=i*salary+i*Hire; intmin;  for(i=2; i<=n;i++){             for(j=people[i];j<=max_people;j++) {min=inf;//with this front there is no need to initialize the DP with O (n^2).                  for(k=people[i-1];k<=max_people;k++)                    if(min>dp[i-1][k]+ (j>=k? ( j*salary+ (j-k) *hire):(j*salary+ (k-j) *Fire)) ) Min=dp[i-1][k]+ (j>=k? ( j*salary+ (j-k) *hire):(j*salary+ (k-j) *Fire )); DP[I][J]=min; }} min=INF;  for(i=people[n];i<=max_people;i++)            if(min>dp[n][i]) min=Dp[n][i]; printf ("%d\n", Min); }    return 0;}

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Employment planning[hdu1158]