[POJ] - 3264 Balanced Lineup [RMQ][ST]

最新推荐文章于 2020-09-21 22:20:26 发布

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本文介绍了一种称为ST算法的方法，用于快速计算指定区间内最大值与最小值之差，适用于处理大规模数据集上的区间查询问题。通过预处理数据结构，该算法能够有效地解决农民约翰游戏中关于奶牛高度差异的问题。

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

题解

这是ST算法，可以对比一下线段树的写法
http://blog.youkuaiyun.com/ctsas/article/details/64955421

#include<cstdio>
#include<cmath>
#include<algorithm>
#define MAX_N 50005
#define INF 0x3f3f3f3f
using namespace std;

int maxsum[MAX_N][19],minsum[MAX_N][19];
void RMQ(int n){
    for(int j=1;j<19;j++)
        for(int i=0;i+(1<<j)<=n;i++){
            maxsum[i][j]=max(maxsum[i][j-1],maxsum[i+(1<<j-1)][j-1]);
            minsum[i][j]=min(minsum[i][j-1],minsum[i+(1<<j-1)][j-1]);
        }
}

int query(int l,int r){
    int k=log(r-l+1.0)/log(2.0);
    return max(maxsum[l][k],maxsum[r-(1<<k)+1][k])-min(minsum[l][k],minsum[r-(1<<k)+1][k]);
}

int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        for(int i=0;i<n;i++){
            scanf("%d",&maxsum[i][0]);
            minsum[i][0]=maxsum[i][0];
        }
        RMQ(n);
        while(m--){
            int l,r;
            scanf("%d%d",&l,&r);
            printf("%d\n",query(l-1,r-1));
        }
    }
    return 0;
}