Balanced Lineup(树状数组 POJ3264)

最新推荐文章于 2021-08-04 15:15:41 发布

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最新推荐文章于 2021-08-04 15:15:41 发布

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原文链接：http://www.cnblogs.com/juechen/p/5255918.html

本文介绍了一种解决区间最值查询问题的有效算法，通过预处理技术实现快速查询，适用于需要频繁进行区间最大值和最小值查询的应用场景。该算法利用预处理数组存储每个区间的最大值和最小值，通过递推方式计算更长区间的最值。

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Balanced Lineup
Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 40493 Accepted: 19035
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

6 3
1
7
3
4
2
5
1 5
4 6
2 2

Sample Output

6
3
0

Source
USACO 2007 January Silver

树状数组模板

#include <map>
#include <set>
#include <queue>
#include <cstring>
#include <string>
#include <cstdio>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

int R[55000][20][2];

int N,Q;

void RMQ()//计算区间的最值(0代表最大值1代表最小值)
{
    for(int j=1;(1<<j)<=N;j++)
    {
        for(int i=0;i+(1<<j)-1<N;i++)
        {
            R[i][j][0]=max(R[i][j-1][0],R[i+(1<<(j-1))][j-1][0]);

            R[i][j][1]=min(R[i][j-1][1],R[i+(1<<(j-1))][j-1][1]);
        }
    }
}

int RMQ_Look(int l,int r)
{
    int ans=0;

    while((1<<(ans+1))<=(r-l+1))
    {
        ans++;
    }

    return max(R[l][ans][0],R[r-(1<<ans)+1][ans][0])-min(R[l][ans][1],R[r-(1<<ans)+1][ans][1]);
}

int main()
{
    int u,v;

    scanf("%d %d",&N,&Q);

    for(int i=0;i<N;i++)
    {
        scanf("%d",&R[i][0][0]);

        R[i][0][1]=R[i][0][0];
    }

    RMQ();

    for(int i=1;i<=Q;i++)
    {
        scanf("%d %d",&u,&v);

        u--;

        v--;

        printf("%d\n",RMQ_Look(u,v));
    }
    return 0;
}