P2949 [USACO09OPEN]工作调度Work Scheduling 题目描述 Farmer John has so very many jobs to do! In order to run the farm efficiently, he must make money on the jobs he does, each one of which takes just one time unit. His work day starts at time 0 and has 1,000…

P2949 [USACO09OPEN]工作调度Work Scheduling 题目描述 Farmer John has so very many jobs to do! In order to run the farm efficiently, he must make money on the jobs he does, each one of which takes just one time unit. His work day starts at time 0 and has 1,000…

题目描述 Farmer John has so very many jobs to do! In order to run the farm efficiently, he must make money on the jobs he does, each one of which takes just one time unit. His work day starts at time 0 and has 1,000,000,000 time units (!). He currently c…

题目描述 约翰有太多的工作要做.为了让农场高效运转,他必须靠他的工作赚钱,每项工作花一个单位时间. 他的工作日从0时刻开始,有10^8个单位时间.在任一时刻,他都可以选择编号1~N的N(1 <= N <= 10^6)项工作中的任意一项工作来完成. 因为他在每个单位时间里只能做一个工作,而每项工作又有一个截止日期,所以他很难有时间完成所有N个工作,虽然还是有可能. 对于第i个工作,有一个截止时间D_i(1 <= D_i <= 10^9),如果他可以完成这个工作,那么他可以获利P_i(…

解题思路 明明一道比较简单的贪心结果挂了好几次23333,就是按照时间排序,然后拿一个小根堆维护放进去的,如果时间允许就入队并且记录答案.如果不允许就从堆里拿一个最小的比较. #include<bits/stdc++.h> using namespace std; ; inline int rd(){ ,f=;char ch=getchar(); :;ch=getchar();} )+(x<<)+ch-';ch=getchar();} return f?x:-x; } int n,…

P2949 [USACO09OPEN]工作调度Work Scheduling 题目标签是单调队列+dp,萌新太弱不会 明显的一道贪心题,考虑排序先做截止时间早的,但我们发现后面可能会出现价值更高却没有时间做的情况 我们需要反悔的操作 于是我们想到用堆,如果当前放不下且当前价值高于已做工作中的最小价值,则删去它加入当前值 类似用堆实现反悔的贪心的经典题目:P1484 种树 #include<queue> #include<cstdio> #include<algorithm&g…

[USACO09OPEN] 工作调度Work Scheduling 题意翻译 约翰有太多的工作要做.为了让农场高效运转,他必须靠他的工作赚钱,每项工作花一个单位时间. 他的工作日从0时刻开始,有10^9个单位时间.在任一时刻,他都可以选择编号1~N的N(1 <= N <= 10^6)项工作中的任意一项工作来完成. 因为他在每个单位时间里只能做一个工作,而每项工作又有一个截止日期,所以他很难有时间完成所有N个工作,虽然还是有可能. 对于第i个工作,有一个截止时间D_i(1 <= D_i &…

题目描述 Farmer John has N cows that need to be milked (1 <= N <= 10,000), each of which takes only one unit of time to milk. Being impatient animals, some cows will refuse to be milked if Farmer John waits too long to milk them. More specifically, cow…

传送门 这个题类似于建筑抢修. 先按照时间排序. 如果当前时间小于任务截止时间就选, 否则,看看当前任务价值是否比已选的任务的最小价值大, 如果是,就替换. 可以用优先队列. ——代码 #include <queue> #include <cstdio> #include <iostream> #include <algorithm> #define LL long long ; int n; LL tim, ans; struct node { LL x,…

洛谷 2953 [USACO09OPEN]牛的数字游戏Cow Digit Game 题目描述 Bessie is playing a number game against Farmer John, and she wants you to help her achieve victory. Game i starts with an integer N_i (1 <= N_i <= 1,000,000). Bessie goes first, and then the two players…