POJ 3264 Balanced Lineup【RMQ問題】
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.
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 3 1 7 3 4 2 5 1 5 4 6 2 2
Sample Output
6 3 0
題解:ST演算法模板題,ST演算法在上一篇文章中。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 50005;
int F[maxn][32];
int G[maxn][32];
int a[maxn], n, q;
int minx = 99999999, maxx = -1;
void ST_prework() {
for(int i = 1; i<= n; i++) F[i][0] = a[i], G[i][0] = a[i];
int t = log(n*1.0) / log(2.0) + 1;
for(int j = 1; j < t; j++) {
for(int i = 1; i <= n - (1<<j) + 1; i++){
F[i][j] = max(F[i][j-1], F[i+(1<<(j-1))][j-1]);
G[i][j] = min(G[i][j-1], G[i+(1<<(j-1))][j-1]);
}
}
}
void ST_query(int l, int r){
int k = log((r-l+1)*1.0) / log(2.0);
maxx = max(F[l][k], F[r-(1<<k)+1][k]);
minx = min(G[l][k], G[r-(1<<k)+1][k]);
}
int main()
{
scanf("%d %d", &n, &q);
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]);
ST_prework();
while(q--){
int l, r;
scanf("%d %d", &l, &r);
ST_query(l, r);
printf("%d\n", maxx - minx);
}
return 0;
}
相關文章
- 【RMQ】poj 3264 Balanced LineupMQ
- POJ3264 Balanced Lineup【RMQ】MQ
- POJ 3264 Balanced Lineup(簡單的RMQ)MQ
- POJ 3264-Balanced Lineup(RMQ-ST演算法)MQ演算法
- POJ 3264-Balanced Lineup詳解(線段樹區間求值)
- POJ 3264 Balanced Lineup 線段樹入門(點的查詢)
- poj--3264Balanced Lineup+ST演算法求區間最大最小值演算法
- B - Gold Balanced Lineup解題報告(張浩盛倫)Go
- poj--2019Cornfields+二維RMQ問題MQ
- $RMQ$問題($ST$表)MQ
- RMQ問題的各種解法MQ
- 資料結構——RMQ(ST表)問題資料結構MQ
- POJ 3368 Frequent values (UVA 11235)(RMQ)MQ
- UVA 11235 經典RMQ問題MQ
- POJ3984-迷宮問題
- poj 1321 棋盤問題 回溯 JavaJava
- POJ 3368-Frequent values(RMQ+離散化-最頻繁的元素)MQ
- RMQMQ
- Balanced Subsequences
- POJ1700 Crossing River 過河問題ROS
- POJ 3279-Fliptile(母牛翻方格-開關問題)
- poj 1182 並查集經典問題並查集
- RMQ模板MQ
- [題解]SP10606 Balanced Numbers
- 淺談RMQMQ
- RMQ求lcaMQ
- HDU Find the hotel(RMQ)MQ
- RMQ演算法MQ演算法
- 線段樹--RMQMQ
- POJ3468 A Simple Problem with Integers---樹狀陣列(區間問題)陣列
- 【轉載】POJ 圖論題目圖論
- 3339: Rmq ProblemMQ
- poj1276 多重揹包問題(二進位制解決方案)
- Leetcode Balanced Binary TreeLeetCode
- [題解]P3059 [USACO12NOV] Concurrently Balanced Strings G
- ACM訓練方案-POJ題目分類ACM
- 【Lintcode】1793. Balanced Sales Array
- Leetcode-Balanced Binary TreeLeetCode