POJ 3368 Frequent values 【ST表RMQ 维护区间频率最大值】

Frequent values

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 23016		Accepted: 8060

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

Source

Ulm Local 2007

题意概括：

给出一串长度为 N 的非递减序列，和 M 次查询，每次查询区间【L，R】的出现频率最高的数的频率。

解题思路：

很妙的一种做法啊！

因为他是非递减数列，所以基于把每一个连续的相同的数作为同一块处理。

那么我们可以把查询区间分成两部分来处理。

第一部分：边界块，区间起点 L 所处的块和区间 R 所处的块。

　　　　　如果两者所处的块相同则说明，该区间在一个块内，区间内的数都相同，那么我们所求的最大频率就是当前区间长度了

　　　　　如果两者所处的块不相同，那么说明除了这两个块还有其他的块，那么剩下的那些块就是接下来讨论的第二部分。

第二部分：很显然剩余的块都是连续的，完整的。那么RMQ维护的区间最大值妥妥的，没毛病（这里的最大值就是预处理的每一块的数出现的频率啦）。

AC code：

 #include <cstdio>

 #include <iostream>

 #include <cstring>

 #include <cmath>

 #define INF 0x3f3f3f3f

 using namespace std;

 const int MAXN = 1e5+;

 int num[MAXN];

 int hh[MAXN];

 int L[MAXN], R[MAXN];

 int b[MAXN];

 int dpmax[MAXN][];

 int mm[MAXN];

 void init_RMQ(int N, int a[])

 {

     mm[] = -;

     for(int i = ; i <= N; i++){

         mm[i] = ((i&(i-)) == )?mm[i-]+:mm[i-];

         dpmax[i][] = b[i];

     }

     for(int ilen = ; ilen <= mm[N]; ilen++){

         for(int i = ; i+(<<ilen)- <= N; i++)

             dpmax[i][ilen] = max(dpmax[i][ilen-], dpmax[i+(<<(ilen-))][ilen-]);

     }

 }

 int get_RMQ(int LL, int RR)

 {

     int k = mm[RR-LL+];

     return max(dpmax[LL][k], dpmax[RR-(<<k)+][k]);

 }

 int main()

 {

     int N, M;

     while(~scanf("%d", &N) && N){

         scanf("%d", &M);

         for(int i = ; i <= N; i++){

             scanf("%d", &num[i]);

         }

         int k = , ll = ;

         memset(b,  ,sizeof(b));

         for(int i = ; i <= N; i++){

             if(i >  && num[i] != num[i-]){

                 for(int j = ll; j < i; j++){

                     L[j] = ll;

                     R[j] = i-;

                 }

                 b[k] = i-ll;

                 ll = i;

                 k++;

             }

             hh[i] = k;

         }

         for(int j = ll; j <= N; j++){

             L[j] = ll;

             R[j] = N;

         }

         b[k] = N-ll+;

         init_RMQ(k, b);

         int U, V;

         while(M--){

             scanf("%d%d", &U, &V);

             int st = hh[U], ed = hh[V];

             if(st == ed){

                 printf("%d\n", V-U+);

                 continue;

             }

             int ans = max(R[U]-U+, V-L[V]+);

             st++, ed--;

             if(st <= ed) ans = max(ans, get_RMQ(st, ed));

             printf("%d\n", ans);

         }

     }

     return ;

 }