（POJ3368）Frequent values RMQ 求区间出现次数最多的数出现的次数

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Frequent values
Description

You are given a sequence of n integers a1 , a2 , … , an 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 ai , … , aj.

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 , … , an (-100000 ≤ ai ≤ 100000, for each i ∈ {1, …, n}) separated by spaces. You can assume that for each i ∈ {1, …, n-1}: ai ≤ ai+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

题意：
给你一个大小为n的非递减整型数组，有q次查询，问[L,R]中出现次数最多的数出现的次数

分析：“来自白书第三章”
由于是非递减的，所以相同的数会在一起出现，我们可以从左到右按数值将数组分段。count[i]表示第i段的数出现的次数，num[p],left[p],right[p]分别表示位置p所在的段的编号和左右端点的位置。
每次查询[L,R]的结果为一下3个部分的最大值：从L到L所在段的结束处的元素的个数即 right[L] - L + 1,从R所在段的开始处到R处的元素的个数即 R - lfet[R] +1 ,中间第num[L]+1段到第num[R]-1段的count的最大值。
特殊情况： 如果L,R在同一段，则答案为R-L+1

AC代码：

// UVa11235 Frequent Values
// Rujia Liu
#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;const int maxn = 100000 + 5;
const int maxlog = 20;// 区间最*大*值
struct RMQ {int d[maxn][maxlog];void init(const vector<int>& A) {int n = A.size();for(int i = 0; i < n; i++) d[i][0] = A[i];for(int j = 1; (1<<j) <= n; j++)for(int i = 0; i + (1<<j) - 1 < n; i++)d[i][j] = max(d[i][j-1], d[i + (1<<(j-1))][j-1]);}int query(int L, int R) {int k = 0;while((1<<(k+1)) <= R-L+1) k++; // 如果2^(k+1)<=R-L+1，那么k还可以加1return max(d[L][k], d[R-(1<<k)+1][k]);}
};int a[maxn], num[maxn], left[maxn], right[maxn];
RMQ rmq;
int main() {int n, q;while(scanf("%d%d", &n, &q) == 2) {for(int i = 0; i < n; i++) scanf("%d", &a[i]);a[n] = a[n-1] + 1; // 哨兵int start = -1;vector<int> count;for(int i = 0; i <= n; i++) {if(i == 0 || a[i] > a[i-1]) { // 新段开始if(i > 0) {count.push_back(i - start);for(int j = start; j < i; j++) {num[j] = count.size() - 1; left[j] = start; right[j] = i-1;}}start = i;}}rmq.init(count);while(q--) {int L, R, ans;scanf("%d%d", &L, &R); L--; R--;if(num[L] == num[R]) ans = R-L+1;else {ans = max(R-left[R]+1, right[L]-L+1);if(num[L]+1 < num[R]) ans = max(ans, rmq.query(num[L]+1, num[R]-1));}printf("%d\n", ans);}}return 0;
}