[poj3264] Balanced Lineup(st表)
来源:互联网 发布:java 网络编程源码 编辑:程序博客网 时间:2024/06/06 20:30
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 with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.
Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.
Input
Line 1: Two space-separated integers, N and Q.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i
Lines N+2..N+Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i
Lines N+2..N+Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.
Output
Lines 1..Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.
Sample Input
6 31734251 54 62 2
Sample Output
630
Source
USACO 2007 January Silver
【题解】RMQ模板题
【代码】
#include<iostream>#include<cstring>#include<cstdio>#include<cmath>#define sz 16using namespace std;int n,m,l,r,Max,Min,k;int a[50005],stmax[50005][sz],stmin[50005][sz];int main(){scanf("%d%d",&n,&m);for (int i=1;i<=n;++i) scanf("%d",&a[i]),stmax[i][0]=stmin[i][0]=a[i];for (int j=1;j<sz;++j) for (int i=1;i<=n;++i) if (i+(1<<j)-1<=n){ stmax[i][j]=max(stmax[i][j-1],stmax[i+(1<<(j-1))][j-1]); stmin[i][j]=min(stmin[i][j-1],stmin[i+(1<<(j-1))][j-1]); }for (int i=1;i<=m;++i){scanf("%d%d",&l,&r);if (l>r) swap(l,r);k=floor(log((double)(r-l+1))/log((double)(2)));Max=max(stmax[l][k],stmax[r-(1<<k)+1][k]);Min=min(stmin[l][k],stmin[r-(1<<k)+1][k]);printf("%d\n",Max-Min);}}
0 0
- [poj3264] Balanced Lineup(st表)
- poj3264 Balanced Lineup(ST表)
- 【poj3264】 Balanced Lineup(st表)
- [学习][poj3264]稀疏表(ST表) Balanced Lineup
- POJ3264 Balanced Lineup 线段树|ST表
- poj3264 Balanced Lineup(RMQ +st)
- POJ3264 Balanced Lineup (RMQ & ST)
- POJ3264 Balanced Lineup(RMQ)
- [POJ 3264] Balanced Lineup (ST表)
- poj3264——Balanced Lineup(ST算法及线段树操作)
- POJ3264 Balanced Lineup 线段树 RMQ ST算法应用
- Poj3264: Balanced Lineup—题解+st表解释
- RMQ (st表) Balanced Lineup
- Balanced Lineup(poj3264,线段树入门)
- POJ3264 Balanced Lineup(线段树入门)
- poj3264 Balanced Lineup(RMQ裸题)
- poj3264 Balanced Lineup(线段树orRMQ)
- poj3264(线段树) Balanced Lineup
- ES6学习——元数据(meta)编程:代理(Proxies)应用示例
- wampsever内置phpmyadmin密码修改和弹窗登陆
- 深入理解Java的接口和抽象类
- 【LCT】BZOJ 2631:tree
- SCOI2015 day1
- [poj3264] Balanced Lineup(st表)
- UVa 796 Critical Links(无向图求割边)
- android ViewHolder模式超简洁写法
- u-boot移植之mmc,网卡配置
- ultraedit 激活办法
- echarts 百度开源图标工具demo
- 请慎用java的File#renameTo(File)方法
- 安卓开发——自定义ViewGroup
- 关于创建spring源码环境时遇到的缺少spring-cglib-repack-3.2.0.jar和spring-objenesis-repack-2.2.jar两个包的问题