建两颗线段树分别存最大和最小值,模板题~ #include<cstdio> #include<algorithm> #include<cstring> using namespace std; ; struct node { int l; int r; int sum; }segTree1[maxn*],segTree2[maxn*]; int a[maxn]; int N; int Q; int x,y; void build1 (int i,int l,int r)…

题目链接:https://vjudge.net/problem/POJ-3264 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 simple, h…

Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 45243   Accepted: 21240 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer Joh…

Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 44720   Accepted: 20995 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer Joh…

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 simple, he will take a contiguous range of cows fr…

Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 32778   Accepted: 15425 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer Joh…

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 simple, he will take a contiguous range of cows fr…

题目传送门 Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 64655   Accepted: 30135 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farm…

http://poj.org/problem?id=3264 Time Limit: 5000MS     Memory Limit: 65536K Description 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 w…

题意:有n头牛,输入他们的身高,求某区间身高的极值的差(max-min), 用RMQ模板,同时构造求极大值和极小值的两个数组. //poj3264 #include <iostream> #include <string> #include <cstring> #include <queue> #include <vector> #include <cstdio> #include<cmath> using namespa…

题目链接:http://poj.org/problem?id=3264 典型RMQ,这道题被我鞭尸了三遍也是醉了…这回用新学的st算法. st算法本身是一个区间dp,利用的性质就是相邻两个区间的最值的最值一定是这两个区间合并后的最值,这条性质决定了这个dp子问题的重叠.可以利用这个性质预处理出这张表,只不过步长是2的幂次. 查询的时候也是如此,但是未必会精准地选中两个区间,不要紧,因为两个区间重叠的部分也会被自动算在求最值的内部.这个时候如果算的是区间和的话,要减去这一部分.(区间和的话直接用前…

题目链接:http://poj.org/problem?id=3264 一排牛按1~n标号记录重量,问每个区间最重的和最轻的差值. 线段树维护当前节点下属叶节点的两个最值,查询后作差即可. #include <algorithm> #include <iostream> #include <iomanip> #include <cstring> #include <climits> #include <complex> #includ…

Q个数 问区间最大值-区间最小值 // #pragma comment(linker, "/STACK:1024000000,1024000000") #include <iostream> #include <cstdio> #include <cstring> #include <sstream> #include <string> #include <algorithm> #include <list&…

本文出自:http://blog.csdn.net/svitter 题意:在1~200,000个数中.取一段区间.然后在区间中找出最大的数和最小的数字.求这两个数字的差. 分析:按区间取值,非常明显使用的线段树. 区间大小取200000 * 4 = 8 * 10 ^5; 进行查询的时候.注意直接推断l, r 与mid的关系就可以.一開始写的时候直接与tree[root].L推断,多余了, 逻辑不对. #include <iostream> #include <stdio.h> #i…

图论 图论解题报告索引 DFS poj1321 - 棋盘问题 poj1416 - Shredding Company poj2676 - Sudoku poj2488 - A Knight's Journey poj1724 - ROADS(邻接表+DFS) BFS poj3278 - Catch That Cow(空间BFS) poj2251 - Dungeon Master(空间BFS) poj3414 - Pots poj1915 - Knight Moves poj3126 - Prim…

Balanced Lineup poj3264 线段树 题意 一串数,求出某个区间的最大值和最小值之间的差 解题思路 使用线段树,来维护最大值和最小值,使用两个查询函数,一个查区间最大值,一个查区间最小值,然后做差就好了,基本上就是线段树模板题 代码实现 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long ll; const int i…

http://poj.org/problem?id=3264 (题目链接) 题意 给出序列,求区间最大值-最小值 Solution 无修改,询问较多,ST表水一发. ST算法(Sparse Table): 它是一种动态规划的方法.以最小值为例.a为所寻找的数组,用一个二维数组 f(i,j) 记录区间 [i,i+2^j-1] 区间中的最小值.其中 f[i,0] = a[i] ; 所以,对于任意的一组 (i,j),f(i,j) = min{ f(i,j-1),f(i+2^(j-1),j-1)} 来使…

Balanced Lineup Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 40493 Accepted: 19035 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John de…

题意: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 <= 180,000) 个可能的牛的选择和所有牛的身高 (1 <= 身高 <= 1,000,000). 他想知道每一组里面最高和最低的牛的身高差别. 注意: 在最大数据上, 输入和输出将占用大部分运…

题意:给你一组值,然后询问某个区间的最大值和最小值得差 分析:因为没有更新,所以只需要查找即可,节点保存一个最大值最小值就行了 ****************************************************************** #include<stdio.h> #include<math.h> #include<; ;} }tree[maxn*]; ].MaxH, tree[root<<|].MaxH);     tree[ro…

http://poj.org/problem?id=3264 题意 rmq max min之差 模板: #define _CRT_SECURE_NO_WARNINGS #include<cmath> #include<iostream> #include<stdio.h> #include<algorithm> using namespace std; #define rep(i,t,n) for(int i =(t);i<=(n);++i) #def…

我早期在csdn的博客之一,正好复习st表就拿过来.http://write.blog.csdn.net/mdeditor#!postId=63713810 这道题其实本身不难(前提是你得掌握线段树或者st表当中的一种) 那么这道题我们来讲一讲st表(因为这题询问次数有点多) 一般关系式 dp[i][j]=min/max(dp[dp[i][j-1],dp[i+pow(2,j-1)][j-1]]) 可以看出来吧,其实这就是动态规划 好的我们来解释一下关系式是什么吧 首先这个是一个二分,具体是什么呢…

题意: 告诉你n头奶牛的高度,然后给你一个区间,你需要求出这个区间最高的奶牛与最矮的奶牛之间相差多少 链接:http://poj.org/problem?id=3264 思路: 线段树区间查询,用两个查询函数,一个查最大值,另一个查最小值,将他们相减即可. 代码: #include <iostream> #include <algorithm> #include <cmath> #include <cstdio> #include <cstdlib&g…

st表很像线段树,但线段树既能查询和修改,而st表只能查询. 首先我们先用二维数组建立一个表,st[i][j]表内存的是从第i位开始1<<j范围内的best(st[i][j-1],st[i+(1<<j-1)][j-1]) 由上面的公式可知st[i][j]是由st[i][j-1],st[i+(1<<j-1)][j-1]更新而来的,我们可以将st[i][j]分成两段,由1<<j=2*(1<<j-1)可得一段是st[i][j-1],另一段则是st[i+…

这里说几篇博客,建议从上到下看 https://blog.csdn.net/qq_31759205/article/details/75008659 https://blog.csdn.net/sgh666666/article/details/80448284 https://www.cnblogs.com/kuangbin/p/3227420.html ----------------------------------------------------------------------…

Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 40687   Accepted: 19137 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer Joh…

poj   3264  Balanced Lineup link: http://poj.org/problem?id=3264 Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 48747   Accepted: 22833 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤…

用户不管是打开A/R Posted Transactions Detail还是A/R Posted Transactions Summary 窗口,均显示如下一个警示: 打开Currency Codes窗口,检查decimal显示格式,均已经设为最大长度,说明与此无关. 后来,找到一个Utility,Modules >> Customer >> General >> Utilities >> Rebalance Customer Balances: Reba…

Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 32820   Accepted: 15447 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer Joh…

总时间限制: 5000ms 内存限制: 65536kB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 Ultimate Frisbee with some of the cows. To keep things simple, he will take…