Balanced Lineup

最新推荐文章于 2022-05-26 15:58:06 发布

转载最新推荐文章于 2022-05-26 15:58:06 发布 · 145 阅读

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原文链接：http://www.cnblogs.com/Yum20/p/9630785.html

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#运维

本文介绍了一种解决区间最大最小值查询问题的有效算法，通过预处理数据建立稀疏表，实现快速查询任意区间内的最大值和最小值，适用于大量查询场景。

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 64964		Accepted: 30260
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

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <climits>
#include <cassert>
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define LL long long
#define MAXX 50000+10
using namespace std;
int maxx[MAXX][32],minn[MAXX][32];
int n,m;
int main(){
    while(~scanf("%d %d",&n,&m)){
        fill(maxx[0],maxx[0]+MAXX*32,0);
        fill(minn[0],minn[0]+MAXX*32,0);
        for(int i=1;i<=n;i++){
            scanf("%d",&minn[i][0]);
        maxx[i][0]    =minn[i][0];
        }
        int z=log(n*1.0)/log(2.0);
        for(int j=1;j<=z;j++){
            for(int i=1;i<=n;i++){
                if(n<(i+(1<<(j-1))) )
                break;
                maxx[i][j]=max(maxx[i][j-1],maxx[i+(1<<(j-1))][j-1]);
                minn[i][j]=min(minn[i][j-1],minn[i+(1<<(j-1))][j-1]); 
            }
        }
        int x,y,sum;
        for(int i=1;i<=m;i++){
            scanf("%d%d", &x, &y);
            int k = log(y - x + 1.0) / log(2.0);
            //int minnn=min(minn[x][k], minn[y - (1<<k) + 1][k]);
    sum=(max(maxx[x][k], maxx[y - (1<<k) + 1][k]) - min(minn[x][k], minn[y - (1<<k) + 1][k]));
    //printf("%d\n",minnn);
            printf("%d\n", sum);
        }
    }
}