poj 3264 st算法_poj st算法-优快云博客

本文介绍了如何利用线段树（Sparse Table）解决给定范围内牛群高度差值的查询问题，包括输入解析、初始化线段树、区间查询等关键步骤。通过实例演示了算法实现过程，并提供了源代码。适用于需要高效处理区间查询场景的开发者。

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 36777		Accepted: 17218
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 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

Sample Output

6
3
0

Source

USACO 2007 January Silver

查询区间最大值和最小值的差。查询最值，用线段树或者sparse table，这里不涉及到更新的操作，而且查询次数Q范围比n大，所以用st做就好了

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>

using namespace std;

#define maxn 50005

int arr[maxn];
int stmax[maxn][17], stmin[maxn][17];
int n, q;

void init()
{
    for(int i = 0; i < n; i++)
        stmax[i][0] = stmin[i][0] = arr[i];
    for(int j = 1; (1<<j)-1 < n; j++){
        for(int i = 0; i+(1<<j)-1 < n; i++){
            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]);
        }
    }
}

int main()
{
    while(~scanf("%d%d", &n, &q)){
        for(int i = 0; i < n; i++)
            scanf("%d", arr+i);

        init();

        int a, b;
        for(int i = 0; i < q; i++){
            scanf("%d%d", &a, &b);
            a--; b--;
            int t = log(b-a+0.0)/log(2.0);
            int maxi = max(stmax[a][t],stmax[b-(1<<t)+1][t]);
            int mini = min(stmin[a][t], stmin[b-(1<<t)+1][t]);
            printf("%d\n", maxi-mini);
        }
    }
    return 0;
}