UVA - 11235 Frequent values

2007/2008 ACM International Collegiate Programming Contest 
University of Ulm Local Contest

Problem F: Frequent values

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 Specification

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 Specification

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

---

A naive algorithm may not run in time!

 

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <algorithm>

 using namespace std;

 const int maxn=;
 int cnt=;
 struct node
 {
     int L,R,nb,v;
 }N[maxn];

 int dp[maxn+][];
 int a[maxn+],kt[maxn+],n,m;

 int RMQ_init()
 {
     for(int i=;i<=cnt;i++) dp[i][]=N[i].nb;
     for(int j=;(<<j)<=cnt;j++)
     {
         for(int i=;i+(<<j)-<=cnt;i++)
         {
             dp[i][j]=max(dp[i][j-],dp[i+(<<(j-))][j-]);
         }
     }
 }

 int RMQ(int L,int R)
 {
     int k=;
     while(R-L+>=(<<k)) k++; k--;
     return max(dp[L][k],dp[R-(<<k)+][k]);
 }

 int main()
 {
     while(scanf("%d",&n)&&n)
     {
         scanf("%d",&m);
         ///yu chu li
         memset(N,,sizeof(N));
         scanf("%d",a+);
         cnt=;N[cnt].v=a[];N[cnt].L=;N[cnt].nb=;
         for(int i=;i<=n;i++)
         {
             scanf("%d",a+i);
             if(a[i]==N[cnt].v)
             {
                 N[cnt].nb++;
             }
             else
             {
                 cnt++;
                 N[cnt-].R=i-;
                 N[cnt].L=i;N[cnt].v=a[i];N[cnt].nb=;
             }
             if(i==n) N[cnt].R=n;
         }
         int pos=;
         kt[]=-;
         for(int i=;i<=cnt;i++)
         {
             int temp=N[i].nb;
             while(temp--)
             {
                 kt[pos]=i;
                 pos++;
             }
         }
         RMQ_init();
         while(m--)
         {
             int a,b,na=-,nb=-;
             scanf("%d%d",&a,&b);
             na=kt[a],nb=kt[b];
             if(na==nb)
             {
                 printf("%d\n",b-a+);
             }
             else if(na+==nb)
             {
                 printf("%d\n",max(N[na].R-a+,b-N[nb].L+));
             }
             else
             {
                 int ans1=N[na].R-a+,ans2=b-N[nb].L+,ans3=;
                 int L=na+,R=nb-;
                 ans3=RMQ(L,R);
                 printf("%d\n",max(ans1,max(ans2,ans3)));
             }
         }
     }
     return ;
 }