POJ 3368 Frequent values(RMQ/线段树区间合并)_线段树合并求区间的k大-优快云博客

题目地址：点击打开链接

题意：给定一个数组，其中的元素满足非递减顺序，要求对于一对起点和终点，回答出其中某个元素重复出现的最大次数。

思路：

由于原数列已经排好序，所以相同的数字一定是在一起的。我们可以O(n)扫描一遍，算出每一个位置上的数字num是从左起第几个num，记为num[i]

如样例，扫描后如下，括号内是num数组值

-1(1) -1(2) 1(1) 1(2) 1(3) 1(4) 3(1) 10(1) 10(2) 10(3)

这样似乎直接求num数组区间[l,r]内的最大值就能得到答案。实际上这样无法得到正确的结果，因为如果查询区间左端点上num值不等于1，就会得到错误的答案。如对样例求[5,7]会得到4。

出现这个错误的原因上面已经说了，因为num数组是对区间[1,n]的计数。

解决方法也很简单，在O(n)倒着扫一遍，对于每个位置上的数字num，求出最右侧的数字为num的位置，记为R[i]，如下

-1(2) -1(2) 1(6) 1(6) 1(6) 1(6) 3(7) 10(10) 10(10) 10(10)

这样查询[l,r]时，只要R[l] < r，就使用RMQ查询[ R[l]+1 , r ]区间内的最大值，这样可以保证查询区间的左端点num值一定是1。而对于[l , R[l] ]这段区间，很显然R[l] - l + 1是唯一一个可能的答案。

当R[l] >= r时，显然[l, r]整个区间都是同一个数字。答案为r-l+1.

这个解法的关键是numl数组记录该位置以左该数的数量，numr数组记录最右和自己相同数到哪。RMQ直接对numl查询（l , r）时左端点的区间可能不

完整，此时就可以用numr数组直接得到在（l , r）中与左端点相同的数数量，然后剩下的范围RMQ查询，取较大值。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int maxn = 1e5+5;
int dp[maxn][25], numl[maxn], numr[maxn], a[maxn], n, m;

void initRMQ()
{
    for(int j = 0; j < 25; j++)
        for(int i = 1; i <= n; i++)
        {
            if(j == 0) { dp[i][j] = numl[i]; continue; }
            int len = 1<<j;
            if(i+len-1 <= n)
                dp[i][j] = max(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
        }
}

int RMQ(int l, int r)
{
    int k = (int)log2(r-l+1);
    return max(dp[l][k], dp[r-(1<<k)+1][k]);
}

int main(void)
{
    while(cin >> n, n)
    {
        cin >> m;
        for(int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        numl[1] = 1;
        for(int i = 2; i <= n; i++)
            numl[i] = a[i]==a[i-1] ? numl[i-1]+1 : 1;
        numr[n] = n;
        for(int i = n-1; i >= 1; i--)
            numr[i] = a[i]==a[i+1] ? numr[i+1] : i;

        initRMQ();
        while(m--)
        {
            int l, r;
            scanf("%d%d", &l, &r);
            if(numr[l] >= r) printf("%d\n", r-l+1);
            else printf("%d\n", max(numr[l]-l+1, RMQ(numr[l]+1, r)));
        }
    }
    return 0;
}

Frequent values

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 18434		Accepted: 6652

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

Ulm Local 2007

POJ 3368 Frequent values(RMQ/线段树区间合并)

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