UVA

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=100100;
int cnt=0;
struct node
{
    int L,R,nb,v;
}N[maxn];

int dp[maxn+10][20];
int a[maxn+10],kt[maxn+10],n,m;

int RMQ_init()
{
    for(int i=1;i<=cnt;i++) dp[i][0]=N[i].nb;
    for(int j=1;(1<<j)<=cnt;j++)
    {
        for(int i=1;i+(1<<j)-1<=cnt;i++)
        {
            dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
        }
    }
}

int RMQ(int L,int R)
{
    int k=0;
    while(R-L+1>=(1<<k)) k++; k--;
    return max(dp[L][k],dp[R-(1<<k)+1][k]);
}



int main()
{
    while(scanf("%d",&n)&&n)
    {
        scanf("%d",&m);
        ///yu chu li
        memset(N,0,sizeof(N));
        scanf("%d",a+1);
        cnt=0;N[cnt].v=a[1];N[cnt].L=1;N[cnt].nb=1;
        for(int i=2;i<=n;i++)
        {
            scanf("%d",a+i);
            if(a[i]==N[cnt].v)
            {
                N[cnt].nb++;
            }
            else
            {
                cnt++;
                N[cnt-1].R=i-1;
                N[cnt].L=i;N[cnt].v=a[i];N[cnt].nb=1;
            }
            if(i==n) N[cnt].R=n;
        }
        int pos=1;
        kt[0]=-1;
        for(int i=0;i<=cnt;i++)
        {
            int temp=N[i].nb;
            while(temp--)
            {
                kt[pos]=i;
                pos++;
            }
        }
        RMQ_init();
        while(m--)
        {
            int a,b,na=-1,nb=-1;
            scanf("%d%d",&a,&b);
            na=kt[a],nb=kt[b];
            if(na==nb)
            {
                printf("%d
",b-a+1);
            }
            else if(na+1==nb)
            {
                printf("%d
",max(N[na].R-a+1,b-N[nb].L+1));
            }
            else
            {
                int ans1=N[na].R-a+1,ans2=b-N[nb].L+1,ans3=0;
                int L=na+1,R=nb-1;
                ans3=RMQ(L,R);
                printf("%d
",max(ans1,max(ans2,ans3)));
            }
        }
    }
    return 0;
}