[RMQ][ST算法]Frequent values

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

思路:预处理第i的数在1~i中出现了多少次，得到freq[i]，这样就转化为了对freq数组的RMQ了，即求max{freq[x~y]}；但其实不是这样的，freq[x]中记录的频率可能要大于a[x]实际在区间[x,y]出现的次数，同理freq[y]也未必准确；
那我们就把区间[x,y]分成两段，一段是[x,t],一段是[t+1,y]；t是与a[x]相等的最后一个元素的下标，那么答案即为max(t-x+1,max{freq[t+1~y]});
AC代码：

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;

int n,q;
int a[100010];
int fre[100010];
int bl[100010];//a[x]所属区域，也可以不要该数组，则r[i]的含义就是a[i]出现的最后位置
int r[100010];//r[i]的含义是区域i的最右下标
int maxn[100010][20];

void ST(){
   for(int i=1;i<=n;i++) maxn[i][0]=fre[i];
   int k=int(log(n*1.0)/log(2.0));
   for(int j=1;j<=k;j++){
     for(int i=1;i<=n;i++){
        if(i+(1<<j)-1>n) break;
        maxn[i][j]=max(maxn[i][j-1],maxn[i+(1<<(j-1))][j-1]);
     }
   }
}

int query(int x,int y){
  int k=int(log(y-x+1.0)/log(2.0));
  return max(maxn[x][k],maxn[y-(1<<k)+1][k]);
}

int main()
{
    while(scanf("%d",&n)!=EOF&&n){
      scanf("%d",&q);
      int cnt=0;
      scanf("%d",&a[1]),fre[1]=1,bl[1]=++cnt;
      for(int i=2;i<=n;i++) {
        scanf("%d",&a[i]);
        if(a[i]==a[i-1]) fre[i]=fre[i-1]+1,bl[i]=bl[i-1],r[bl[i]]=i;
        else fre[i]=1,bl[i]=++cnt,r[bl[i]]=i;
      }
      ST();
      while(q--){
        int x,y;
        scanf("%d%d",&x,&y);
        if(bl[x]==bl[y]) printf("%d
",y-x+1);
        else printf("%d
",max(r[bl[x]]-x+1,query(r[bl[x]]+1,y)));
      }
    }
    return 0;
}