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

Source

Ulm Local 2007

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+,INF=1e9;

inline int read(){

    char c=getchar();int x=,f=;

    while(c<''||c>''){if(c=='-')f=-;c=getchar();}

    while(c>=''&&c<=''){x=x*+c-'';c=getchar();}

    return x*f;

}

int n,q,l,r;

int x,last,cnt=,a[N],v[N],c[N],id[N],left[N],right[N];

int st[N][];

void initRMQ(){

    memset(st,,sizeof(st));

    for(int i=;i<=cnt;i++) st[i][]=c[i];

    for(int j=;(<<j)<=cnt;j++)

        for(int i=;i+(<<j)-<=cnt;i++)

            st[i][j]=max(st[i][j-],st[i+(<<(j-))][j-]);

}

int rmq(int l,int r){

    if(l>r) return ;

    int k=log(r-l+)/log();

    return max(st[l][k],st[r-(<<k)+][k]);

}

int main(int argc, const char * argv[]) {

    while((n=read())){

        q=read();

        memset(c,,sizeof(c));

        memset(left,,sizeof(left));

        memset(right,,sizeof(right));

        v[]=v[n+]=INF;

        for(int i=;i<=n;i++){

            v[i]=read();

            if(v[i]==v[i-]){

                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=;i<=q;i++){

            l=read();r=read();

            int ans=;

            if(id[l]==id[r]) ans=r-l+;

            else{

                ans=max(right[id[l]]-l+,r-left[id[r]]+);

                ans=max(ans,rmq(id[l]+,id[r]-));

            }

            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 ;

}

巴特西

POJ3368Frequent values[RMQ 游程编码]

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