POJ3262贪心

【POJ3262贪心】的更多相关文章

POJ-3262 贪心的一个小技巧

Protecting the Flowers Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 3204 Accepted: 1300 Description Farmer John went to cut some wood and left N (2 ≤ N ≤ 100,000) cows eating the grass, as usual. When he returned, he found to his ho…

题意:FJ去砍树,然后和平时一样留了 N (2 ≤ N ≤ 100,000)头牛吃草.当他回来的时候,他发现奶牛们正在津津有味地吃着FJ种的美丽的花!为了减少后续伤害,FJ决定立即采取行动:运输每头牛回到自己的牛棚. 每只奶牛i在离牛棚Ti(1 ≤ Ti ≤ 2,000,000) 分钟路程的地方,每分钟吃掉Di(1 ≤ Di ≤ 100)朵花.FJ只能一次带回一头奶牛.弄回一头奶牛i来回需要2*Ti分钟.从带回i号奶牛开始,i号奶牛就不会吃花.请你找出被毁坏的花的最小数量 . 题解:按照排序贪心…

【贪心算法】POJ-3262

一.题目 Description Farmer John went to cut some wood and left N (2 ≤ N ≤ 100,000) cows eating the grass, as usual. When he returned, he found to his horror that the cluster of cows was in his garden eating his beautiful flowers. Wanting to minimize the…

POJ3262 Protecting the Flowers 【贪心】

Protecting the Flowers Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 4418 Accepted: 1785 Description Farmer John went to cut some wood and left N (2 ≤ N ≤ 100,000) cows eating the grass, as usual. When he returned, he found to his ho…

《挑战程序设计竞赛》2.2 贪心法-其它 POJ3617 3069 3253 2393 1017 3040 1862 3262

POJ3617 Best Cow Line 题意给定长度为N的字符串S,要构造一个长度为N的字符串T.起初,T是一个空串,随后反复进行下列任意操作: 从S的头部(或尾部)删除一个字符,加到T的尾部目标是构造字典序尽可能小的字符串T. 思路贪心算法,不断取S的开头和末尾中较小的一个字符放到T的末尾.但对于S的开头和末尾字符相同的情况下,需要比较下一个字符大小,这可以用如下算法实现: 按照字典序比较S和S翻转后的字符串S1,如果S较小,则从S的开头取,否则从末尾取. 代码 Source Cod…

POJ_3262 Protecting the Flowers 【贪心】

一.题面 POJ3262 二.分析这题要往贪心上面想应该还是很容易的,但问题是要证明为什么比值关系就能满足. 可以选择几个去分析,入1-6 与 2-15 和 1-6 与2-5 和 1-6 与 2- 12. 三.AC代码 #include <cstdio> #include <iostream> #include <algorithm> #include <fstream> using namespace std; ; struct Cow { int…

【POJ - 3262】Protecting the Flowers（贪心）

Protecting the Flowers 直接中文 Descriptions FJ去砍树,然后和平时一样留了 N (2 ≤ N ≤ 100,000)头牛吃草.当他回来的时候,他发现奶牛们正在津津有味地吃着FJ种的美丽的花!为了减少后续伤害,FJ决定立即采取行动:运输每头牛回到自己的牛棚. 每只奶牛i在离牛棚Ti(1 ≤ Ti ≤ 2,000,000) 分钟路程的地方,每分钟吃掉Di(1 ≤ Di ≤ 100)朵花.FJ使尽浑身解数,也只能一次带回一头奶牛.弄回一头奶牛i需要2*Ti分钟(来回…