HDU1158 动态规划
Employment Planning
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1486 Accepted Submission(s): 555
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
一种枚举状态的 DP, 设在了 i 天前有 j 个员工的最小花费是 dp[i][j], 那么可以枚举 i 和 j, 通过转移方程 :dp[i][j]=min(dp[i-1][k]+s[k][j]+pay*k)(a[i-1]<=k<=max(a)) 求出每个状态的最优解 , 然后把 dp[n][j] 加上 j*pay 就是最终的花费 , 暴力找出最小值即可 .
代码如下 :
#include<stdio.h> #include<string.h> #define inf 2000000000 int p,f,h; int tran(int i,int j) { if (i>j) return (i-j)*f; if (i<j) return (j-i)*h; return 0; } inline int small (int a,int b) { return a<b?a:b; } int main() { int dp[15][10000]; int i,j,k,n,max,min,res,s,t; int a[15]; while (scanf("%d",&n)&&n) { scanf("%d%d%d",&h,&p,&f); max=0; min=inf; for (i=1; i<=n; ++i) { scanf("%d",&a[i]); if (a[i]>max) max=a[i]; if (a[i]<min) min=a[i]; } for (i=1; i<=n; ++i) { for (j=min; j<=max; ++j) { dp[i][j]=inf; dp[0][j]=0; } } for (i=1; i<=n; ++i) { for (j=min; j<=max; ++j) { if (i==1) s=t=0; else { s=a[i-1]; t=max; } for (k=s; k<=t; ++k) { dp[i][j]=small(dp[i][j],dp[i-1][k]+tran(k,j)+k*p); } } } res=inf; for (j=a[n];j<=max;++j) {dp[n][j]+=j*p;if (dp[n][j]<res) res=dp[n][j];} printf("%d/n",res); } return 0; }