POJ 3368—— Frequent values（频繁出现的数值UVA11235） RMQ

Frequent values

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 13288		Accepted: 4880

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

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#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<cstdlib>
#include<cmath>
#define M 100000+10
using namespace std;
int n,m;
int a[M],num[M],lef[M],rig[M],dp[M][20];

vector<int> cnt;

void RMQ_init(const vector<int> &A)
{
    int n=A.size();
    for(int i=0;i<n;++i) dp[i][0]=A[i];
    for(int j=1;(1<<j)<=n;++j){
        for(int i=0;i+(1<<j)-1<n;++i){
            dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
        }
    }
}
void RLE()
{
    int l=-1;
    cnt.clear();
    for(int i=0;i<=n;++i){
        if(i==0||a[i]>a[i-1]){
            if(i>0){
                cnt.push_back(i-l);
                for(int j=l;j<i;++j){
                    num[j]=cnt.size()-1;
                    lef[j]=l;
                    rig[j]=i-1;
                }
            }
            l=i;
        }
    }
    RMQ_init(cnt);

}
int RMQ(int l,int r)
{
    int k = 0;
	while((1<<(k+1)) <= r-l+1) k++;
    return max(dp[l][k],dp[r-(1<<k)+1][k]);
}
int main()
{
    while(~scanf("%d",&n)&&n){
        scanf("%d",&m);
        for(int i=0;i<n;++i) scanf("%d",&a[i]);
        a
=a[n-1]+1;//?
        RLE();
        while(m--){
            int l,r,ans;
            scanf("%d %d",&l,&r);
            l--,r--;
            if(num[l]==num[r]) ans=r-l+1;
            else{
                ans=max(r-lef[r]+1,rig[l]-l+1);
                if(num[l]+1<=num[r]-1) ans=max(ans,RMQ(num[l]+1,num[r]-1));
            }
            printf("%d\n",ans);
        }
    }
    return 0;
}