POJ 3368 Frequent values（RMQ）

Frequent values

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15134		Accepted: 5519

Description

You are given a sequence of n integers a₁ , a₂ , ... , a_n in non-decreasing order. In addition to that, you are given several queries consisting of indices i and j (1 ≤ i ≤ j ≤ n). For each query, determine the most frequent value among the integers a_i , ... , a_j.

Input

The input consists of several test cases. Each test case starts with a line containing two integers n and q (1 ≤ n, q ≤ 100000). The next line contains n integers a₁ , ... , a_n (-100000 ≤ a_i ≤ 100000, for each i ∈ {1, ..., n}) separated by spaces. You can assume that for each i ∈ {1, ..., n-1}: a_i ≤ a_i+1. The following q lines contain one query each, consisting of two integers i and j (1 ≤ i ≤ j ≤ n), which indicate the boundary indices for the 
query.

The last test case is followed by a line containing a single 0.

Output

For each query, print one line with one integer: The number of occurrences of the most frequent value within the given range.

Sample Input

10 3
-1 -1 1 1 1 1 3 10 10 10
2 3
1 10
5 10
0

Sample Output

1
4
3

Source

   题意：给出一个长度为n的不下降序列，然后m次查询,每次输入两个数l,r，输出区间[l,r]中连续的数中最长的数目是多少。

   思路：将输入的不下降序列进行预处理。找出当中的相邻的数字的相邻的数目。存储在p数组中，然后作为RMQ算法中dp的初值，每次输入l,r时。输出该区间的最大值就能够了。

   ps:只是之中有一个须要注意的地方，就是每次询问[l,r]时，要先将这个区间分成两个部分，前半部分求出与num[pt] == num[l]的个数，直到num[pt]!=num[l]，当然pt<=r然后对[pt,r]进行RMQ求解，这样做的原因是排除在l之前还有数与num[l]相等。

这种话可能会影响结果。然后将RMQ求解得到的结果与pt-l的值进行取最大值就能够了。

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>

using namespace std;

const int N = 200010;
int maxx[N][20];
int num[N];
int p[N];
int n,m;

void play() {
    int l = floor(log10(double(n))/log10(double(2)));
    for(int j=1; j<=l; j++) {
        for(int i=1; i<=n+1-(1<<j); i++) {
            maxx[i][j] = max(maxx[i][j-1],maxx[i+(1<<(j-1))][j-1]);
        }
    }
}

int RMQ(int xx,int yy) {
    if(xx>yy) {
        return 0;
    }
    int pp = floor(log10(double(yy-xx))/log10(double(2)));
    return max(maxx[xx][pp],maxx[yy-(1<<pp)+1][pp]);
}

int main() {
    while(scanf("%d",&n)!=EOF) {
        if(n == 0) {
            break;
        }
        scanf("%d",&m);
        for(int i=1; i<=n; i++) {
            scanf("%d",&num[i]);
            if(i == 1) {
                p[i] = 1;
            } else {
                if(num[i] == num[i-1]) {
                    p[i] = p[i-1] + 1;
                } else {
                    p[i] = 1;
                }
            }
        }
        for(int i=1; i<=n; i++) {
            maxx[i][0] = p[i];
        }
        play();
        int x,y;
        while(m--) {
            scanf("%d%d",&x,&y);
            int pt = x;
            while(pt<=y && num[pt] == num[pt-1]) {
                pt++;
            }
            int ans = RMQ(pt,y);
            ans = max(ans,pt-x);
            printf("%d\n",ans);
        }

    }
    return 0;
}