Employment Planning Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5292 Accepted Submission(s): 2262 Problem DescriptionA project manager wants to determine the number of the workers needed in eve…

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…

Employment Planning Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5151    Accepted Submission(s): 2208 Problem Description A project manager wants to determine the number of the workers neede…

Employment Planning Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1158 Description A project manager wants to determine the number of the workers needed in every month. He does know the minima…

题目传送门 Employment Planning Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6242    Accepted Submission(s): 2710 Problem Description A project manager wants to determine the number of the workers…

Employment Planning 有n个月,每个月有一个最小需要的工人数量\(a_i\),雇佣一个工人的费用为\(h\),开除一个工人的费用为\(f\),薪水为\(s\),询问满足这n个月正常工作的最小费用,\(n\leq 12\). 解 显然可以猜一个结论,因为工人数不确定,猜测每一个月的工人数量必然为某一个月的工人的最小数量,于是可以设\(f[i][j]\)表示前i个月,拥有工人数量\(b_j\)的最小费用,其中\(b\)为\(a\)的离散化数组,因此有 \(f[i][j]=min(m…

Employment Planning Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7097    Accepted Submission(s): 3034 Problem Description A project manager wants to determine the number of the workers needed…

又一次看题解. 万事开头难,我想DP也是这样的. 呵呵,不过还是有进步的. 比如说我一开始也是打算用dp[i][j]表示第i个月份雇j个员工的最低花费,不过后面的思路就完全错了.. 不过这里还有个问题,这样开数组j开多大比较好,难道要我开2^31-1这么大? 题解里面开了1000多,也许再小一点也能过吧. 因为有可能解雇一个人的花费比较大,所以某个月可能继续雇佣他这样总的算来是最省的. 所以第i个月可能雇佣的人数是从num[i] ~ NumMax. 首先对第一个月的费用初始化,就是(雇佣+薪水)…

题意:给出n个月,雇佣一个人所需的钱hire,一个人工作一个月所需要的钱salary,解雇一个人所需要的钱fire,再给出这n个月每月1至少有num[i]个人完成工作,问完成整个工作所花费的最少的钱是多少. 用dp[i][j]表示在第i个月雇佣j个人所需要的最少花费 先考虑只解雇人和聘请人的情况 for(j=num[i];j<=sum;j++) { ])//说明雇佣了人 dp[i][j]=dp[i-][num[i-]]+j*salary+(j-num[i-])*hire; else//说明解聘了…

题目链接 题意 : n个月,每个月都至少需要mon[i]个人来工作,然后每次雇佣工人需要给一部分钱,每个人每个月还要给工资,如果解雇人还需要给一笔钱,所以问你主管应该怎么雇佣或解雇工人才能使总花销最小. 样例解释 : 3个月,雇佣一个工人需要4钱,每个工人每个月的工资是5钱,解雇一个工人需要6钱.然后这三个月每个月需要的工人数最少分别是10人9人11人.0结束输入 3 4 5 6 10 9 11 0 主管在一开始就招10个人并且第二个月不解雇,因为解雇比他工资贵,需要花费40+50+50 = 1…