题意:给定1-m的区间,然后给定n个小区间,用最少的小区间去覆盖1-m的区间,覆盖不了,输出-1. 析:一看就知道是贪心算法的区间覆盖,主要贪心策略是把左端点排序,如果左端点大于1无解,然后, 忽略小于1的部分(如果有的话),再找最长的区间,然后把这个区间的右端点作为下次寻找的起点, 再找最大区间,直到覆盖到最后. 注意:首先要判断好能不能覆盖,不能覆盖就结束,有可能会提前结束,也要做好判断,我就在这WA了好几次, 悲剧...其他的就比较简单了,不用说了. 代码如下: #include <ios…

Cleaning Shifts Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some cleaning chores around the barn. He always wants to have one cow working on cleaning things up and has divided the day into T shifts (1 <= T <= 1,000,000…

<pre name="code" class="html"> Cleaning Shifts Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14425   Accepted: 3700 Description Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some cleani…

Cleaning Shifts 题目连接: http://poj.org/problem?id=2376 Description Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some cleaning chores around the barn. He always wants to have one cow working on cleaning things up and has divided…

POJ 2376 题意: 给出一给大区间和n各小区间,问最少可以用多少小区间覆盖整个大区间. 分析: 贪心法.设t为当前所有已确定区间的最右端,那我们可以每次都取所有可选的小区间(左端点<=t+1)中右端点最大的值,然后更新最右端点ans++.初始时t=0 注:所谓衔接不是[0,1][1,2]这样首尾相接,而是[0,1][2,3]即可,故为 t+1 #include<iostream> #include<algorithm> #include<string.h>…

POJ 2376 Cleaning Shifts(轮班打扫) Time Limit: 1000MS   Memory Limit: 65536K [Description] [题目描述] Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some cleaning chores around the barn. He always wants to have one cow working on clean…

POJ 3190 Stall Reservations贪心 Description Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be milked over some precise time interval A..B (1 <= A <= B <= 1,000,000), which includes both times A and B. Obvi…

POJ 2392 Space Elevator(贪心+多重背包) http://poj.org/problem?id=2392 题意: 题意:给定n种积木.每种积木都有一个高度h[i],一个数量num[i].另一个限制条件,这个积木所在的位置不能高于limit[i],问能叠起的最大高度? 分析: 本题是一道多重背包问题, 只是每一个物品的选择不只要受该种物品的数量num[i]限制, 且该物品还受到limit[i]的限制. 这里有一个贪心的结论: 我们每次背包选取物品时都应该优先放置当前limit…

题目链接:http://poj.org/problem?id=2376 题目大意:选择一些区间使得能够覆盖1-T中的每一个点,并且区间数最少 题目分析:这道题目很明显可以用贪心法来解决.但题目没有看起来那么简单,有许多的坑. 我的贪心策略如下: 1.将区间按照起点排序,并且保证起点相同的,终点大的排在前边 2.在前一个选取的区间范围[L0,R0+1]中,选取起点在此范围但终点最靠右的一个区间 3.重复这个过程 另外,还有几点需要注意的地方: 1.要保证第一个区间起点和最后一个区间终点符合1-L的…

题目:http://poj.org/problem?id=2376 题意:就是 N 个区间, 输入 N 个区间的 [begin, end],求能用它们覆盖区间[1,T]的最小组合. 题解: 1. 首先对所有奶牛的排序,按照开始时间升序排序. 2. 更新 起点 为 上一次的终点 + 1,并寻找覆盖起点,且终点最远的区间 #include <iostream> #include <cstdlib> #include <cstdio> #include <algorit…