[USACO08FEB]酒店Hotel 线段树 题面 其实就是区间多维护一个lmax,rmax(表示从左开始有连续lmax个空房,一直有连续rmax个空房到最右边),合并时讨论一下即可. void push_up(int x, int l, int r){ int mid=(l+r)>>1; if(sum[lson]==mid-l+1) lmax[x]=sum[lson]+lmax[rson]; else lmax[x]=lmax[lson]; if(sum[rson]==r-(mid+1)+…

Hotel 转载自:http://www.cnblogs.com/scau20110726/archive/2013/05/07/3065418.html [题目链接]Hotel [题目类型]线段树 区间合并 &题意: 有一个线段,从1到n,下面m个操作,操作分两个类型,以1开头的是查询操作,以2开头的是更新操作 1 w 表示在总区间内查询一个长度为w的可用区间,并且要最靠左,能找到的话返回这个区间的左端点并占用了这个区间,找不到返回0 好像n=10 , 1 3 查到的最左的长度为3的可用区间就…

http://poj.org/problem?id=3667 Hotel Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 9484   Accepted: 4066 Description The cows are journeying north to Thunder Bay in Canada to gain cultural enrichment and enjoy a vacation on the sunny s…

Hotel The cows are journeying north to Thunder Bay in Canada to gain cultural enrichment and enjoy a vacation on the sunny shores of Lake Superior. Bessie, ever the competent travel agent, has named the Bullmoose Hotel on famed Cumberland Street as t…

Hotel Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 14958   Accepted: 6450 Description The cows are journeying north to Thunder Bay in Canada to gain cultural enrichment and enjoy a vacation on the sunny shores of Lake Superior. Bessie…

题目传送门 酒店 题目描述 The cows are journeying north to Thunder Bay in Canada to gain cultural enrichment and enjoy a vacation on the sunny shores of Lake Superior. Bessie, ever the competent travel agent, has named the Bullmoose Hotel on famed Cumberland Str…

题意:n个空房间.两种操作:1.选择最小的连续D个房间入住,并输出这连续D个房间的最小标号.2.将某个区间内的房间全部退房. #include <cstdio> #include <iostream> #include <algorithm> #define ll long long #define lson l, m, rt<<1 #define rson m+1, r, rt<<1|1 #define st first #define nd…

题目链接 线段树的区间合并. 和上一题差不多....第三种操作只需要输出maxx[1]的值就可以. #include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <map> #include <set> #include <s…

题目连接:http://poj.org/problem?id=3667 题意:1 a:询问是不是有连续长度为a的空房间,有的话住进最左边 2 a b:将[a,a+b-1]的房间清空 线段树操作:update:区间替换 query:询问满足条件的最左断点 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #define LL long long #define M…

经典线段树的题. 每个节点存储的信息:左端点连续空房间的长度,右端点连续空房间长度,连续空房间的最大长度. 由于要求每次必须从尽量靠左边的位置进行居住,那么搜索时应尽量让区间起始位置更小: 1.如果当前区间的最大空房间长度小于要寻找的长度,说明不会找到符合的区间,直接退出. 2.如果左子区间的最大空房间长度大于等于要寻找的长度,那么应该进入左子区间查找. 3.否则,如果左子区间的右端点连续空房间长度 + 右子区间的左端点连续空房间的长度 大于等于 要寻找的长度则直接返回左子区间右端点起始空房间下…