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…

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=<<…

<题目链接> 题目大意: 给定一段序列,进行q次询问,输出每次询问区间的最大值与最小值之差. 解题分析: RMQ模板题,用ST表求解,ST表用了倍增的原理. #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; typedef long long ll; ; int n,q; ll maxsum[M][],mi…

打出st表的步骤:1:建立初始状态,2:区间按2的幂从小到大求出值 3:查询时按块查找即可 #include<iostream> #include<cstring> #include<cstdio> using namespace std; #define maxn 50010 ],mi[maxn][],d[maxn]; void initmax(int n,int d[]){ ;i<=n;i++) mx[i][]=d[i]; ;(<<j)<=n…

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…

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…

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,…

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…

发现lca的倍增解法和st表差不多..原理都是一样的 /* 整篇文章分成两部分,中间没有图片的部分,中间有图片的部分 分别用ST表求f1,f2表示以第i个单词开始,连续1<<j行能写多少单词 */ #include<bits/stdc++.h> #define FIN freopen("in.txt","r",stdin); using namespace std; #define ll long long #define MX 100005…

ST表,稀疏表,用于求解经典的RMQ问题.即区间最值问题. Problem: 给定n个数和q个询问,对于给定的每个询问有l,r,求区间[l,r]的最大值.. Solution: 主要思想是倍增和区间dp. 状态:dp[i][j] 为闭区间[i,i+2^j-1]的最值. 这个状态与转移方程的关系很大,即闭区间的范围涉及到了转移方程的简便性. 转移方程:dp[i][j]=max(dp[i][j-1],dp[i+2^(j-1)][j-1]). 这是显然的,但这里有个细节:第一个项的范围为[i,i+2^…