POJ3368Frequent values[RMQ 游程编码]

Frequent values

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17581		Accepted: 6346

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

Source

Ulm Local 2007

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调试到00:30

白书上的

把相同的RLE，cnt段，a是数值，c出现次数，left和right是这一段左右到原来位置那里，id[p]是p位置的编号

用RMQ快速求id[l]+1到id[r]-1段的最大值，其他的直接加减就行了

//
//  main.cpp
//  poj3368
//
//  Created by Candy on 10/8/16.
//  Copyright © 2016 Candy. All rights reserved.
//

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const int N=1e5+5,INF=1e9;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
int n,q,l,r;
int x,last,cnt=0,a[N],v[N],c[N],id[N],left[N],right[N];
int st[N][20];
void initRMQ(){
    memset(st,0,sizeof(st));
    for(int i=1;i<=cnt;i++) st[i][0]=c[i];
    for(int j=1;(1<<j)<=cnt;j++)
        for(int i=1;i+(1<<j)-1<=cnt;i++)
            st[i][j]=max(st[i][j-1],st[i+(1<<(j-1))][j-1]);
}
int rmq(int l,int r){
    if(l>r) return 0;
    int k=log(r-l+1)/log(2);
    return max(st[l][k],st[r-(1<<k)+1][k]);
}
int main(int argc, const char * argv[]) {
    while((n=read())){
        q=read();
        memset(c,0,sizeof(c));
        memset(left,0,sizeof(left));
        memset(right,0,sizeof(right));
        v[0]=v[n+1]=INF;
        for(int i=1;i<=n;i++){
            v[i]=read();
            if(v[i]==v[i-1]){
                c[cnt]++;
                right[cnt]=i;
                id[i]=cnt;
            }else{
                cnt++;
                a[cnt]=v[i];
                c[cnt]++;
                left[cnt]=right[cnt]=i;
                id[i]=cnt;
            }
        }
        initRMQ();
        //for(int i=1;i<=cnt;i++) printf("init %d %d %d %d\n",a[i],c[i],left[i],right[i]);
        for(int i=1;i<=q;i++){
            l=read();r=read();
            int ans=0;
            if(id[l]==id[r]) ans=r-l+1;
            else{
                ans=max(right[id[l]]-l+1,r-left[id[r]]+1);
                ans=max(ans,rmq(id[l]+1,id[r]-1));
            }
            printf("%d\n",ans);
        }
    }
    //printf("\n\n\n%d %d %d %d",id[1]+1,c[2],id[10]-1,rmq(id[1]+1,id[10]-1));
    return 0;
}