[HDU 1806] Frequent values

Frequent values

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1146    Accepted Submission(s): 415

Problem Description

You are given a sequence of n integers a ₁ , a ₂ , ... , 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 a ₁ , ... , 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

 

Hint

A naive algorithm may not run in time!

 

分成三段、中间RMQ、然后求最大值即可

#include <iostream>

#include <cstdio>

#include <cmath>

using namespace std;

#define N 100010



int n,m;

int a[N];

int dp[N][20];

int id[N];

int len[N],l[N],r[N];



void init()

{

    int i,j;

    for(i=1;i<=n;i++)

    {

        dp[i][0]=len[i];

    }

    int k=(int)(log((double)n)/log(2.0));

    for(j=1;j<=k;j++)

    {

        for(i=1;i+(1<<j)-1<=n;i++)

        {

            dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);

        }

    }

}

int query(int i,int j)

{

    int k=(int)(log((double)(j-i+1))/log(2.0));

    int res=max(dp[i][k],dp[j-(1<<k)+1][k]);

    return res;

}

int main()

{

    int i,pos;

    while(scanf("%d",&n),n)

    {

        pos=0;

        scanf("%d",&m);

        for(i=1;i<=n;i++)

        {

            scanf("%d",&a[i]);

            if(a[i]!=a[i-1])

            {

                pos++;

                l[pos]=i;

            }

            id[i]=pos;

            r[pos]=i;

            len[pos]=r[pos]-l[pos]+1;

        }

        n=pos;

        init();

        int x,y,xx,yy,ans1,ans2,ans3;

        while(m--)

        {

            int x,y;

            scanf("%d%d",&x,&y);

            xx=id[x];

            yy=id[y];

            if(xx==yy) ans1=ans2=y-x+1; //特殊情况、当x和y在同一个区间、答案是y-x+1

            else

            {

                ans1=r[xx]-x+1;

                ans2=y-l[yy]+1;

            }

            ans3=0;

            if(xx+1<=yy-1)ans3=query(xx+1,yy-1);

            printf("%d\n",max(max(ans1,ans2),ans3));

        }

    }

    return 0;

}