【随便做做|线段树】POJ3667 Hotel

题意：1→n1\to n的房间，初始为空。有两种操作：
- 操作11: 需要选出连续一段的若干个房间入住，输出起点位置，选不出则输出00。
- 操作22: 将从leftleft开始的连续sizesize个房间清空。
- 数据规模: n,m≤50000n,m\le 50000
题解：用线段树|区间树维护。节点主要维护三个值:{ml,mr,md}:\{ml,mr,md\}，分别表示从左端开始连续，从右端开始向左连续，整段区间连续的最长子区段的长度。对第一种操作，只需要无脑queryquery，然后对queryquery到的结果([from,to][from,to])进行更新即可；对第二种操作，当然就和上述更新差不多辣~
P.S. 为了保证按照上述方法维护时线段树的单次操作复杂度，需要加入lazylazy操作，呃。。。胡乱地搞一下，就可以通过了。。。

/* **********************************************File Name: 3667.cppAuther: zhengdongjian@tju.edu.cnCreated Time: 2015/9/22 星期二 下午 9:16:53*********************************************** */
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <complex>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <sstream>
#include <utility>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;typedef long long ll;
typedef pair<int, int> P;const int MAX = 50007;
struct Node {int ml;int mr;int md;int lazy;
} room[MAX << 2];void build(int root, int left, int right) {room[root].ml = room[root].mr = room[root].md = right - left + 1;room[root].lazy = 0;if (left == right) {return;}int mid = (left + right) >> 1;build(root << 1, left, mid);build(root << 1 | 1, mid + 1, right);
}void pushup(int root, int left, int right) {int ls = root << 1;int rs = root << 1 | 1;int mid = (left + right) >> 1;room[root].ml = room[ls].ml;if (room[ls].ml == mid - left + 1) {room[root].ml += room[rs].ml;}room[root].mr = room[rs].mr;if (room[rs].mr == right - mid) {room[root].mr += room[ls].mr;}room[root].md = max(max(room[ls].md, room[rs].md), max(room[root].ml, room[root].mr));if (room[ls].mr + room[rs].ml > room[root].md) {room[root].md = room[ls].mr + room[rs].ml;}
}void pushdown(int root, int left, int right) {//printf("pushdown (%d, %d)\n", left, right);if (left == right) {room[root].ml = room[root].mr = room[root].md = (~room[root].lazy) ? 0 : 1;room[root].lazy = 0;return;}room[root << 1].lazy = room[root << 1 | 1].lazy = room[root].lazy;if (~room[root].lazy) {room[root << 1].ml = room[root << 1].mr = room[root << 1].md = 0;room[root << 1 | 1].ml = room[root << 1 | 1].mr = room[root << 1 | 1].md = 0;} else {int mid = (left + right) >> 1;room[root << 1].ml = room[root << 1].mr = room[root << 1].md = mid - left + 1;room[root << 1 | 1].ml = room[root << 1 | 1].mr = room[root << 1 | 1].md = right - mid;}room[root].lazy = 0;
}void check_out(int root, int left, int right, int from, int to) {//printf("check out (%d, %d) with %d, %d...", left, right, from, to);if (to < left || from > right) {//puts("not match");return;} else if (from <= left && to >= right) {//puts("all match");room[root].ml = room[root].mr = room[root].md = right - left + 1; //all clearroom[root].lazy = -1;return;}//puts("match some");if (room[root].lazy != 0) {pushdown(root, left, right);}int mid = (left + right) >> 1;check_out(root << 1, left, mid, from, to);check_out(root << 1 | 1, mid + 1, right, from, to);pushup(root, left, right);
}int query(int root, int left, int right, int cnt) {//printf("query %d, %d, %d (%d, %d, %d)...", left, right, cnt, room[root].ml, room[root].mr, room[root].md);if (room[root].md < cnt) {//printf("(%d, %d, %d) not enough!\n", room[root].ml, room[root].mr, room[root].md);return 0;}if (left == right) {//printf("equal to one\n");return left;}//puts("to judge:");if (room[root].lazy != 0) {pushdown(root, left, right);}int mid = (left + right) >> 1;int le = query(root << 1, left, mid, cnt);if (le > 0) {//printf("get left\n");return le;} if (room[root << 1].mr + room[root << 1 | 1].ml >= cnt) {//printf("get mid\n");return mid - room[root << 1].mr + 1;}//printf("get right\n");return query(root << 1 | 1, mid + 1, right, cnt);
}void check_in(int root, int left, int right, int from, int to) {//printf("check in %d, %d, with %d, %d...", left, right, from, to);if (to < left || from > right) {//printf("match failed!\n");return;} else if (from <= left && to >= right) {//puts("match all!");room[root].lazy = 1;room[root].ml = room[root].mr = room[root].md = 0;return;}//puts("match some");if (room[root].lazy) {pushdown(root, left, right);}int mid = (left + right) >> 1;check_in(root << 1, left, mid, from, to);check_in(root << 1 | 1, mid + 1, right, from, to);pushup(root, left, right);//printf("and update (%d, %d) to %d, %d, %d\n", left, right, room[root].ml, room[root].mr, room[root].md);
}void debug(int root, int left, int right) {printf("(%d, %d) => (%d, %d, %d, %d)\n", left, right, room[root].ml, room[root].mr, room[root].md, room[root].lazy);if (left == right) {return;}int mid = (left + right) >> 1;debug(root << 1, left, mid);debug(root << 1 | 1, mid + 1, right);
}int main() {int n, m;while (~scanf(" %d %d", &n, &m)) {build(1, 1, n);int op, sz, le;while (m--) {scanf(" %d", &op);if (op == 1) {scanf(" %d", &sz);le = query(1, 1, n, sz);//printf("le = %d\n", le);if (le) check_in(1, 1, n, le, le + sz - 1);printf("%d\n", le);} else {scanf(" %d %d", &le, &sz);check_out(1, 1, n, le, le + sz - 1);}//debug(1, 1, n);}}return 0;
}

【随便做做|线段树】POJ3667 Hotel相关推荐

和Leo一起做爱线段树的好孩子HDU5238 Calculator
额神仙题感觉就是一个暴力啊线段树维护的是一个类似于键值线段树的东西? 每个叶子节点才表示一个操作而在PushUp的时候合并答案这个咋合并? 暴力合并.我们记录下当前模意义下所有的答案然后暴 ...
和Leo一起做爱线段树的好孩子之火车运输
ByteLand火车站(编号0)每天都要发往全国各地N列客运火车,编号1 N.第i列火车的目的地是编号Si的火车站. 对任意车站X,都与X+1车站有铁轨直接相连,因此火车站可以看成数轴上的整数点,第i ...
线段树学习心得 poj3468
2016.5.10 神奇的博主回来了今天,当然讲的就是加强版的堆,其实算线段树的说... (还有你们看过的留留言好不好呀,谢谢了) --------------------------- 今天说的就 ...
poj 2352 Stars 线段树(先建后查/边建边查)/树状数组三种方法思路详解,带你深入了解线段树难度⭐⭐⭐★
poj 2352 Stars 目录 poj 2352 Stars 1.树状数组 2.线段树,先建树后查找 3.线段树,边建树边查找 Description Astronomers often exam ...
Master of GCD（差分数组||线段树）
题意:长度为n的数组,一开始都是1.对于区间操作l,r,x,在l~r上乘以x.x2||x3.问操作完毕之后,n个数的最大公因子是多少. 对于每个x,都等于2或者是3.那么看最大的公因子,就看各个位置上 ...
P7988-[USACO21DEC] HILO G【set,线段树】
正题题目链接:https://www.luogu.com.cn/problem/P7988 题目大意给出一个长度为nnn的排列,开始有一个数字xxx,第一次询问回答x<a1x<a_1x ...
POJ 2528 Mayor's posters (离散化和线段树使用)
还是做了线段树的题目,玩了两天也要继续看看题目了.之前就有看离散化的概念,大家可以去百度百科一下,简单转载一个例子离散化的大概思路 : 比如说给你一组数据 1 4 1000 100000, 如果 ...
C++线段树运用进阶：【高级数据结构】自动售票系统
前言线段树是什么我就不在这里详细讲解了大家在用线段树时,那些简单的题型应该都没有问题了吧? --那好,我们一起来看一道稍微复杂一点的水题吧! [高级数据结构]自动售票系统题目描述某次列车途径C个 ...
【XSY3367】青春野狼不做姐控偶像的梦（线段树）
题意:给一个 1∼n1\sim n1∼n 的排列,多次询问某段区间内的值域连续子区间的个数. 区间值域连续的另一种表达方式:max−min=r−lmax-min=r-lmax−min=r−l,即 (m ...
【做练习】K-th Number（线段树）线段树的原理
1. 题目描述 You are working for Macrohard company in data structures department. After failing your pre ...

【随便做做|线段树】POJ3667 Hotel

【随便做做|线段树】POJ3667 Hotel相关推荐

最新文章

热门文章