Cleaning Shifts Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 9   Accepted Submission(s) : 2 Problem Description Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some clea…

这是个典型的线程服务区间模型.一些程序要在一段时间区间上使用一段线程运行,问至少要使用多少线程来为这些程序服务? 把所有程序以左端点为第一关键字,右端点为第二关键字从小到大排序.从左向右扫描.处理当前区间时,提取出所有线程中最后一个被服务中的区间中右端点最小的区间(可用小根堆实现),若当前区间左端点值大于提取出的区间的右端点的值,则把当前区间安排到选中的区间的那个线程,否则只能再派出一个线程来负责该区间了. 此贪心是正确的,因为正在被服务中的区间中右端点最小的区间,能使当前区间不被该线程负责的当…

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…

[USACO06FEB] Stall Reservations 贪心 \(n\)头牛,每头牛占用时间区间\([l_i,r_i]\),一个牛棚每个时间点只能被一头牛占用,问最少新建多少个牛棚,并且每头牛在哪个牛棚里? 比较巧的\(O(n)\)扫一遍做法,用一个小跟堆维护所有牛棚最后一个牛占用的时间(即\(r_i\)),贪心的想,如果最先结束的那个牛棚都不能满足当前牛,那么我们只能新开一个牛棚,并继续维护小根堆:如果那个牛棚满足,那么这个牛就去那个牛棚,更新当前牛棚最后占用时间. #include…

这里 3190 Stall Reservations 按照吃草时间排序 之后我们用 优先队列维护一个结束时间 每次比较堆顶 看是否满足 满足更新后放到里面不满足就在后面添加 #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> using namespace std; ; int n,c[maxn]; struct ac{…

Stall Reservations Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3106   Accepted: 1117   Special Judge 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 in…

Stall Reservations 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. Obviously, FJ must create a…

https://vjudge.net/problem/POJ-3190 cin和scanf差这么多么..tle和300ms 思路:先对结构体x升序y升序,再对优先队列重载<,按y升序. 然后依次入队,如果node[i].x<=q.top().y ans++, 否则出队,入队,使用出队的那个摊位. #include<iostream> #include<cstdio> #include<queue> #include<cstring> #inclu…

题意:给定N头奶牛,每头牛有固定的时间[a,b]让农夫去挤牛奶,农夫也只能在对应区间对指定奶牛进行挤奶, 求最少要多少个奶牛棚,使得在每个棚内的奶牛的挤奶时间不冲突. 思路:1.第一个想法就是贪心,对每头牛的挤奶时间[a,b]按a和b都从小排序,接着从左边开始找地一头牛, 然后再往右边找能够不冲突的牛再一个奶牛棚内.这个算法事件复杂度为n*n,由于最多5000头牛 所以后面还是TLE了. 2.还是自己太弱了,原来可以用优先队列进行优化,可以把当前冲突的牛放入优先队列,然后每次都能够冲优先队列 里…

我一开始想用线段树,但是发现还要记录每头cow所在的棚...... 无奈之下选择正解:贪心. 用priority_queue来维护所有牛棚中结束时间最早的那个牛棚,即可得出答案. 注意代码实现的细节. #include <cstdio> #include <algorithm> #include <queue> using namespace std; ; struct Cow { int a, b, num; bool operator < (const Cow…

题意 (来自洛谷) 约翰的N(l<N< 50000)头奶牛实在是太难伺候了,她们甚至有自己独特的产奶时段.当 然对于某一头奶牛,她每天的产奶时段是固定的,为时间段A到B包括时间段A和时间段B.显然,约翰必须开发一个调控系统来决定每头奶牛应该被安排到哪个牛棚去挤 奶,因为奶牛们显然不希望在挤奶时被其它奶牛看见. 约翰希望你帮他计算一下:如果要满足奶牛们的要求,并且每天每头奶牛都要被挤过奶,至少需要多少牛棚 •每头牛应该在哪个牛棚被挤奶.如果有多种答案,你只需任意一种即可. 分析 一道比较经典的贪…

Stall Reservations 原文是English,这里直接上中文吧 Descriptions: 这里有N只 (1 <= N <= 50,000) 挑剔的奶牛! 他们如此挑剔以致于必须在[A,B ]的时间内产奶(1 <= A <= B <= 1,000,000)当然, FJ必须为他们创造一个决定挤奶时间的系统.当然,没有牛想与其他奶牛分享这一时光 帮助FJ做以下事: 使每只牛都有专属时间的最小牛棚数 每只牛在哪个牛棚 也许有很多可行解.输出一种即可,采用SPJ Inp…

Stall Reservations Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11002   Accepted: 3886   Special Judge 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 i…

[USACO06FEB]摊位预订Stall Reservations 题目描述 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. Obviou…

Stall Reservations Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4434   Accepted: 1588   Special Judge 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 in…

Stall Reservations Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7646   Accepted: 2710   Special Judge 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 in…

P2859 [USACO06FEB]摊位预订Stall Reservations 题目描述 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.…

http://poj.org/problem?id=3190 Stall Reservations Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3567   Accepted: 1276   Special Judge Description Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be m…

                                                                                            Stall Reservations Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5469   Accepted: 1973   Special Judge Description Oh those picky N (1 <= N <=…

1651: [Usaco2006 Feb]Stall Reservations 专用牛棚 Time Limit: 10 Sec  Memory Limit: 64 MBSubmit: 509  Solved: 280[Submit][Status] Description Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be milked over some precise…

线段树.. -------------------------------------------------------------------------------------- #include<cstdio> #include<cstring> #include<algorithm> #include<iostream>   #define rep( i , n ) for( int i = 0 ; i < n ; i++ ) #define…

题目 1651: [Usaco2006 Feb]Stall Reservations 专用牛棚 Time Limit: 10 Sec  Memory Limit: 64 MBSubmit: 553  Solved: 307[Submit][Status] Description Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be milked over some preci…

1651: [Usaco2006 Feb]Stall Reservations 专用牛棚 Time Limit: 10 Sec  Memory Limit: 64 MBSubmit: 566  Solved: 314[Submit][Status] Description Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be milked over some precise…

P2859 [USACO06FEB]摊位预订Stall Reservations 维护一个按右端点从小到大的优先队列 蓝后把数据按左端点从小到大排序,顺序枚举. 每次把比右端点比枚举线段左端点小的数据从优先队列中删掉. 在整个过程中队列的最大长度即为答案. 总之用优先队列模拟一下就ok了 对于luogu需要输出方案数的问题: 再开一个优先队列存未用的编号 每次有线段进队时取走最小的编号,出队时再还回来. 似乎暴力也行(大雾) #include<iostream> #include<cst…

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. Obviously, FJ must create a reservation system…

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. Obviously, FJ must create a reserv…

题目链接: https://vjudge.net/problem/POJ-3190 题目大意: 有N头奶牛,每头奶牛都会在[1,1000000]的时间区间内的子区间进行挤奶.挤奶的时候奶牛一定要单独放在一个牛棚中.一头奶牛的结束时间与另一头奶牛的开始时间重合的时候2头奶牛不能放在同一个牛棚中,例A牛时间[5,10],B牛时间[10,12],那么需要用2个牛棚放这2头牛.输入N头牛的所有的挤奶时间区间,要求输出最少要建几个牛棚,以及每头牛挤奶的时候在第几号牛棚. 思路: 将奶牛根据挤奶开始时间排序…

http://poj.org/problem?id=3190 有n头挑剔的奶牛,只会在一个精确时间挤奶,而一头奶牛需要占用一个畜栏,并且不会和其他奶牛分享,每头奶牛都会有一个开始时间和结束时间,问至少需要多少个 畜栏  并且输出奶牛 i 在哪个畜栏 内挤奶. 首先应该对奶牛以开始时间从小到大排序,然后每次在开始的奶牛中选择结束时间最小的奶牛,这就需要用优先队列.看是否可以用这个畜栏,可以就加入到队列中,不可以就重新加一个畜栏. #include <iostream> #include <…

POJ 3190 题意: 一些奶牛要在指定的时间内挤牛奶,而一个机器只能同时对一个奶牛工作.给你每头奶牛的指定时间的区间(闭区间),问你最小需要多少机器.思路:先按奶牛要求的时间起始点进行从小到大排序,然后维护一个优先队列,里面以已经开始挤奶的奶牛的结束时间早为优先.然后每次只需要检查当前是否有奶牛的挤奶工作已经完成的机器即可,若有,则换那台机器进行工作.若没有,则加一台新的机器.注意:利用priority_queue<Node,vector<Node>,less<Node>…

这个题方法还挺多的,不过洛谷上要输出方案所以用堆最方便 先按起始时间从小到大排序. 我用的是greater重定义优先队列(小根堆).用pair存牛棚用完时间(first)和牛棚编号(second),每次查看队首的first是否比当前牛的起始时间早,是则弹出队首记录当前牛的答案,再把新的pair放进去,否则增加牛棚,同样要塞进队里 #include<iostream> #include<cstdio> #include<algorithm> #include<que…