传送门:http://poj.org/problem?id=3368 Frequent values Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 23016   Accepted: 8060 Description You are given a sequence of n integers a1 , a2 , ... , an in non-decreasing order. In addition to that,…

1. 概述 RMQ(Range Minimum/Maximum Query),即区间最值查询,是指这样一个问题:对于长度为n的数列A,回答若干询问RMQ(A,i,j)(i,j<=n),返回数列A中下标在i,j之间的最小/大值.这两个问题是在实际应用中经常遇到的问题,下面介绍一下解决这两种问题的比较高效的算法.当然,该问题也可以用线段树(也叫区间树)解决,算法复杂度为:O(N)~O(logN),这里我们暂不介绍. 2.RMQ算法 对于该问题,最容易想到的解决方案是遍历,复杂度是O(n).但当数据量…

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=<<…

Summer again! Flynn is ready for another tour around. Since the tour would take three or more days, it is important to find a hotel that meets for a reasonable price and gets as near as possible! But there are so many of them! Flynn gets tired to loo…

/*RMQ 更新最小值操作 By:draymonder*/ #include <iostream> #include <cstdio> using namespace std; ; << ; const int INF = 0x3f3f3f3f; typedef long long LL; *maxn-]; void init(int a)//把元素个数扩充到2的幂 简便 { n = ; while (n < a) n*=; ;i<*n-;i++) s[i]…

类似hdu5289,但是二分更复杂.本题枚举左端点,右端点是一个区间,需要二分找到区间的左端点和右端点(自己手动模拟一次),然后区间长度就是结果增加的次数 另外结果开long long 保存 /** 二分法,枚举左端点,向右寻找第一个最大值不等于最小值的端点 */ #include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #include<cmath> us…

  Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 36864   Accepted: 17263 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 J…

二维RMQ其实和一维差不太多,但是dp时要用四维 /* 二维rmq */ #include<iostream> #include<cstring> #include<cstdio> #include<cmath> using namespace std; #define maxn 305 int val[maxn][maxn],n,m; ][]; void ST(){ ;i<=n;i++) ;j<=m;j++) dpmax[i][j][][]=v…

#include<iostream> #include<cmath> #include<cstdio> #define N 50005 using namespace std; ],minn[N][],a[N]; int ST(int n) { ;i<=n;i++) { maxx[i][]=a[i]; minn[i][]=a[i]; } ;(<<j)<=n;j++) { ;i+(<<j)-<=n;i++) { maxx[i][j…

Max answer https://nanti.jisuanke.com/t/38228 Alice has a magic array. She suggests that the value of a interval is equal to the sum of the values in the interval, multiplied by the smallest value in the interval. Now she is planning to find the max…