浅谈\(RMQ\):https://www.cnblogs.com/AKMer/p/10128219.html 题目传送门:https://lydsy.com/JudgeOnline/problem.php?id=1636 题目传送门:https://lydsy.com/JudgeOnline/problem.php?id=1699 裸的\(RMQ\) 时间复杂度:\(O(nlogn+m)\) 空间复杂度:\(O(nlogn)\) 代码如下: #include <cstdio> #includ…

1699: [Usaco2007 Jan]Balanced Lineup排队 Description 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).…

1699: [Usaco2007 Jan]Balanced Lineup排队 Time Limit: 5 Sec  Memory Limit: 64 MB Description 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…

RMQ.. ------------------------------------------------------------------------------- #include<cstdio> #include<cstring> #include<algorithm> #include<iostream>   #define rep( i , n ) for( int i = 0 ; i < n ; ++i ) #define clr( x…

题面:P2880 [USACO07JAN]平衡的阵容Balanced Lineup 题解: ST表板子 代码: #include<cstdio> #include<cstring> #include<iostream> #define max(a,b) ((a)>(b)?(a):(b)) #define min(a,b) ((a)<(b)?(a):(b)) using namespace std; ,max_log=,maxlog=,inf=<<…

Gold Balanced Lineup Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13540   Accepted: 3941 Description Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared…

http://www.lydsy.com/JudgeOnline/problem.php?id=1699 我是用树状数组做的..rmq的st的话我就不敲了.. #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <algorithm> using namespace std; #defi…

[算法]线段树 #include<cstdio> #include<cctype> #include<algorithm> using namespace std; ; ]; int n,q,a[maxn]; int read_t;char c; int read() { read_t=; while(!isdigit(c=getchar())); +c-';}while(isdigit(c=getchar())); return read_t; } void buil…

要求区间取min和max,可以用st表或线段树维护 st表 #include<iostream> #include<cstdio> using namespace std; const int N=100005; int n,q,b[N],mn[N][20],mx[N][20]; int read() { int r=0,f=1; char p=getchar(); while(p>'9'||p<'0') { if(p=='-') f=-1; p=getchar();…

题意:链接 方法:线段树 解析: 题意即题解. 多次询问区间最大值与最小值的差.显然直接上线段树或者rmq维护区间最值就可以. 代码: #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define N 50010 #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #de…