UVA 1513 - Movie collection option=com_onlinejudge&Itemid=8&page=show_problem&category=502&problem=4259&mosmsg=Submission+received+with+ID+13965699" target="_blank" style="">题目链接 题意:有一些光盘,一開始是n-1叠上去的(1最顶),如今…

题意:给你n盘歌碟按照(1....n)从上到下放,接着m个询问,每一次拿出x碟,输出x上方有多少碟并将此碟放到开头 直接想其实就是一线段的区间更新,单点求值,但是根据题意我们可以这样想 首先我们倒着存  n--1,接着每次询问时把放碟子放到最后,这样我们要开一个映射数组映射每个碟子在哪个位置 其中我们需要使用树状数组维护每个绝对位置是否有碟子(有些碟子已经放到了后面了),再使用区间求和就好了 #include<set> #include<map> #include<queue…

#include<stdio.h> #include<string.h> #include<stdlib.h> #define N 200010 #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 ],tag[N/]; int n,q,cnt; void PushUp(int rt) { sum[rt]=sum[rt<<]+sum[rt<<|]; } void build(i…

题意:有n个影碟,标号为1~n,位置为0~n-1,每次取出一个影碟看完后,将其放在最前面(标号为0处),问每个影碟取出前,其位置之前有多少个影碟. 分析: 1.数组大小开为100000*2,后100000个位置放初始的n个影碟,每次看完一个影碟后,依次向前放在前100000个位置中. 2.将放置影碟处标记为1---add(x, 1),当影碟取出时,再add(x,-1). 3.通过sum(x)-1可知当前位置之前有多少个影碟. #include<cstdio> #include<cstri…

题目链接:1513 - Movie collection 题意:有一堆电影,按1-n顺序排,有m次操作,每次询问第ai个电影之前有多少个电影,然后将其抽出放在堆顶. 分析:线段树应用. 因为每次查询后要将电影移至堆顶,所以我们可以将线段树的区间开到maxn+n,[1,maxn]先置0,在[maxn+1,maxn+n]建树将值置为1,线段树每一段区间的值代表该区间的和. 每次查询后要将该位置的值更新为0,由于要将电影移至堆顶,用一个指针cnt初始化为maxn,记录当前电影可以移到的位置, 由于位置…

UVA - 11987 Almost Union-Find I hope you know the beautiful Union-Find structure. In this problem, you’re to implement something similar, but not identical. The data structure you need to write is also a collection of disjoint sets, supporting 3 oper…

Cellular Network Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 1456 A cellular network is a radio network made up of a number of cells each served by a base station located in the cell. The base sta…

10798 - Be wary of Roses You've always been proud of your prize rose garden. However, some jealous fellow gardeners will stop at nothing to gain an edge over you. They have kidnapped, blindfolded, and handcuffed you, and dumped you right in the middl…

题目链接:UVA 811 Description Once upon a time, in a faraway land, there lived a king. This king owned a small collection of rare and valuable trees, which had been gathered by his ancestors on their travels. To protect his trees from thieves, the king or…

1.Collection是所有集合的父类,在JDK1.5之后又加入了Iterable超级类(可以不用了解) 2.学习集合从Collection开始,所有集合都继承了他的方法 集合结构如图:…