[POJ1275]Cashier Employment 题意: 超市经历已经提供一天里每一小时需要出纳员的最少数量————R(0),R(1),...,R(23).R(0)表示从午夜到凌晨1:00所需要出纳员的最少数目:R(1)表示凌晨1:00到2:00之间需要的:等等.每一天,这些数据都是相同的. 有N人申请这项工作,每个申请者i在每天24小时当中,从一个特定的时刻开始连续工作恰好8小时.定义ti(0<=ti<=23)为上面提到的开始时刻,也就是说,如果第i个申请者被录用,他(或她)将从ti时…

题目: Description A supermarket in Tehran is open 24 hours a day every day and needs a number of cashiers to fit its need. The supermarket manager has hired you to help him, solve his problem. The problem is that the supermarket needs different number…

[POJ 1275] Cashier Employment(差分约束系统的建立和求解) Cashier Employment Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7569   Accepted: 2856 Description A supermarket in Tehran is open 24 hours a day every day and needs a number of cashiers to f…

[题目链接] 点击打开链接 [算法] 设Ti为第i小时有多少个出纳员开始工作,Vi表示第i小时有多少个来应聘的出纳员 那么,有 : 1. 0 <= Ti <= Vi 2. Ti + Ti-1 + Ti-2 + Ti-3 + Ti-4 + Ti-5 + Ti-6 + Ti-7 >= Ri 令Si = T1 + T2 + T3 + ... Ti 则 : 1. Si >= Si-1 2. Si - Si-1 <= Vi 3. Si >= Si-8 + Ri( 8 <=…

简单DP. #include <cstdio> #include <cstring> #include <cstdlib> #include <climits> #define MAXN 105 ][MAXN]; ]; int getMin(int a, int b) { return a<b?a:b; } int main() { int n, max, min; int i, j, k, r, tmp, ans; int hire, salary,…

Cashier Employment Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7997   Accepted: 3054 Description A supermarket in Tehran is open 24 hours a day every day and needs a number of cashiers to fit its need. The supermarket manager has hir…

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http://acm.hdu.edu.cn/showproblem.php?pid=1529 Cashier Employment Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1489    Accepted Submission(s): 672 Problem Description A supermarket in Tehran…

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