Employment Planning
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4782 Accepted Submission(s): 2019
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
参考大牛的题解。
dp[i][j]表示前i个月最后一个月的总人数为j所花的最小费用
状态移动方程:dp[i][j] = min{dp[i-1][k] + cost[i][j]},其中cost[i][j]是第i月的花费,
1~当k<=j时，第i个月请了人所以cost[i][j] = j*salary + (j-k)*hire
2~当k>j时，第i个月炒了人，所以cost[i][j] = j*salary + (k-j)*fire
输入时记录了最多需要的人数。
因为他给的是每个月最少需要的人数，所以for(该月需要的最少人数——max_people)而不是for(1——该月需要的最少人数)
直接初始化第一个月。
状态移动方程:dp[i][j] = min{dp[i-1][k] + cost[i][j]},其中cost[i][j]是第i月的花费,
1~当k<=j时，第i个月请了人所以cost[i][j] = j*salary + (j-k)*hire
2~当k>j时，第i个月炒了人，所以cost[i][j] = j*salary + (k-j)*fire
输入时记录了最多需要的人数。
因为他给的是每个月最少需要的人数，所以for(该月需要的最少人数——max_people)而不是for(1——该月需要的最少人数)
直接初始化第一个月。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 const int INF = 0x3f3f3f3f; 7 int dp[15][100]; 8 int people[15]; 9 void solve(){ 10 int n; 11 int hire,salary,fire; 12 while(scanf("%d",&n) == 1&&n){ 13 scanf("%d%d%d",&hire,&salary,&fire); 14 int max_people = 0; 15 for(int i = 1; i<=n; i++){ 16 scanf("%d",&people[i]); 17 max_people = max(max_people,people[i]); 18 } 19 // memset(dp,INF,sizeof(dp)); 20 for(int i = people[1]; i<=max_people; i++){ 21 dp[1][i] = (hire+salary)*i; 22 } 23 for(int i = 2; i<=n; i++){ 24 for(int j = people[i]; j<=max_people; j++){ 25 int minn = INF; 26 for(int k = people[i-1]; k<=max_people; k++){ 27 int cost = k<=j?(salary*j+(j-k)*hire):(salary*j+(k-j)*fire); 28 minn =min(minn,(dp[i-1][k] + cost)); 29 } 30 dp[i][j] = minn; 31 } 32 } 33 int ans = INF; 34 for(int j = people[n]; j<=max_people; j++) ans = min(ans,dp[n][j]); 35 printf("%d ",ans); 36 } 37 } 38 int main() 39 { 40 solve(); 41 return 0; 42 }