lineup专题
POJ3264 Balanced Lineup 线段树|ST表
Balanced Lineup Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 39453 Accepted: 18511Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000)
poj3264--Balanced Lineup(RMQ求最大最小)
Balanced Lineup Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 33665 Accepted: 15830Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000)
poj3274--Gold Balanced Lineup(hash)
Gold Balanced Lineup Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 12334 Accepted: 3618 Description Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has
![[英语单词] lineup](https://img-blog.csdnimg.cn/direct/f29890d92cc148a683d3054f1979320e.jpeg)
Balanced Lineup （poj-3264）
RMQ解决方案： #include <iostream>#include <math.h>#define max(a,b) ((a>b)?a:b)#define min(a,b) (a<b?a:b)using namespace std;const int maxn=50001;int h[maxn];int mx[maxn][16],mn[maxn][16];int
POJ 3264 Balanced Lineup RMQ / 线段树
题意：给出一个队列，找出指定区间的最大值与最小值。 题解：RMQ, 注意边界需要理清楚。 参考http://www.cnblogs.com/cnjy/archive/2009/08/30/1556566.html RMQ(Range Minimum/Maximum Query)问题： RMQ问题是求给定区间中的最值问题。当然，最简单的算法是O(n)的，但是对于查询次数很多（设
Balanced Lineup POJ-3264（线段树）
For the daily milking, Farmer John’s N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things
poj Balanced Lineup
http://poj.org/problem?id=3264 线段树板子题，由原来的维护区间和变成维护区间最大值和最小值即可。 #include<iostream>using namespace std;typedef long long ll;#define maxn 50007 #define ls l,m,rt<<1#define rs m+1,r,rt<<1|1int m
POJ 3264 Balanced Lineup (线段树单点更新 区间查询)
Balanced Lineup Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 36820 Accepted: 17244Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 5
POJ 3264 Balanced Lineup (RMQ模板)
Balanced Lineup Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 65283 Accepted: 30409Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) alway
POJ 3264 Balanced Lineup 线段树
A - Balanced Lineup For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of t
poj 3264——Balanced Lineup
线段树 #include<iostream>#include<cstdio>using namespace std;#define maxn 50004#define ls (rt<<1)#define rs (rt<<1|1)#define mid ((t[rt].l+t[rt].r)>>1)struct tree{int l,r;int max,min;}t[maxn<<2
[BZOJ1699][Usaco2007 Jan]Balanced Lineup排队
[Usaco2007 Jan]Balanced Lineup排队 时间限制: 1 Sec 内存限制: 128 MB 题目描述 每天,农夫 John 的N(1 <= N <= 50,000)头牛总是按同一序列排队. 有一天, John 决定让一些牛们玩一场飞盘比赛. 他准备找一群在对列中为置连续的牛来进行比赛. 但是为了避免水平悬殊,牛的身高不应该相差太大. John 准备了Q (1 <= Q
【算法】Balanced Lineup
RMP算法的基本应用。 #include<bits/stdc++.h> using namespace std; #define MAXN 200005 int cow[MAXN],sm[MAXN][20],bg[MAXN][20]; int N,Q,len,l,r; void read() { scanf("%d%d",&N,&Q); for (int i=1;i<=N;i++
Balanced Lineup POJ - 3264（RMQ）
Balanced Lineup POJ - 3264 题目连接 题意：给出一个数列，Q个询问，问区间[A, B]中最大值与最小值的差； 思路：线段树可以做，维护最大最小值，直接查找就可以；但是现在要用RMQ做； 何为RMQ？(Range Minimum/Maximum Query) 区间最值询问，通过O(nlogn)的预处理可以在O(1)的时间内找到区间的最值；下面以最大值为例： 令Fm
P2880 [USACO07JAN] Balanced Lineup G 题解
文章目录 题目描述输入格式输出格式样例样例输入样例输出 数据范围与提示完整代码 题目描述 For the daily milking, Farmer John’s N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of
POJ-3264 Balanced Lineup(RMQ)
Balanced Lineup Time Limit: 5000MSMemory Limit: 65536KTotal Submissions: 73096Accepted: 33626Case Time Limit: 2000MS Description For the daily milking, Farmer John’s N cows (1 ≤ N ≤ 50,000) alway
Pku oj 3264 Balanced Lineup（RMQ）
Balanced Lineup Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 46162 Accepted: 21674Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000)
BZOJ 1637: Balanced Lineup 巧妙变换
Description Farmer John 决定给他的奶牛们照一张合影，他让 N (1 ≤ N ≤ 50,000) 头奶牛站成一条直线，每头牛都有它的 坐标(范围: 0..1,000,000,000)和种族(0或1)。 一直以来 Farmer John 总是喜欢做一些非凡的事，当然这次照相 也不例外。他只给一部分牛照相，并且这一组牛的阵容必须是“平衡的”。平衡的阵容，指的是在一组牛
POJ 3264 Balanced Lineup 线段树 / 平方分割
一、题目大意 给出一个长度为 n（n<=50000) 数组 arr，进行Q次查询（Q<=200000），每次查询的内容为数组arr在 [L , R] 的切片的极差（最大元素 - 最小元素） 二、解题思路 1、线段树 区间极差其实就是 区间内最大值 - 区间内最小，那么就想到RMQ，用线段树去维护一个区间内的最大和最小元素，然后根据问题的区间 L 和 R，找到相关的线段树节点，从中找出 最大