hdu 1806 RMQ

Frequent values

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1459    Accepted Submission(s): 535

Problem 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

Hint

A naive algorithm may not run in time!

题意：

查找区间中出现频率最高的，并输出出现次数。

思路：

开始想了很久不知道怎么用到RMQ的，总感觉和最值没什么关系，后来发现对连续的数编号然后求出最大值就好了。

只是需要判断一下不是从1开始的，所以用的二分查找。

但是强行被二分Gank一波。总感觉二分用的不怎么熟练，好久去研究一下。就因为设置的方向不一样，写起来特别麻烦，导致后来还是要重写。

假设我想取一个最接近左边的值，所以设定的l = mid，但是就会导致2和3的时候卡死。后来换了下方向就行了

当你取到3的时候，下一次还是会往下跳，- -！基础真的糟

/*
*/

#include <functional>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <Map>
using namespace std;
typedef long long ll;
typedef long double ld;

using namespace std;

const int maxn = 100010;

int dp[maxn][20];
int m[maxn];
int a[maxn];
int b[maxn];

void iniRMQ(int n,int c[])
{
    m[0] = -1;
    for(int i = 1; i <= n; i++)
    {
        m[i] = ((i&(i-1)) == 0)? m[i-1]+1:m[i-1];
        dp[i][0] = c[i];
    }
    for(int j = 1; j <= m[n]; j++)
    {
        for(int i = 1; i+(1<<j)-1 <= n; i++)
            dp[i][j] = max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
    }
}

int RMQ(int x,int y)
{
    int k = m[y-x+1];
    return max(dp[x][k],dp[y-(1<<k)+1][k]);
}

int get_bi(int s,int t)
{
    int l =s,r = t;
    int to = a[t];
    while(l < r)
    {
        int mid = ((l+r)>>1);
        if(a[mid] >= to) r = mid;
        else l = mid+1;
    }
    return r;
}

int main()
{
    int n,m,T;
    while(scanf("%d",&n) != EOF && n)
    {
        scanf("%d",&m);
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&a[i]);
        }
        int t;
        for(int i = n; i >= 1; i--)
        {
            if(i == n)
                t = 1;
            else
            {
                if(a[i] == a[i+1])t++;
                else t=1;
            }
            b[i] = t;
        }
//        for(int i = 1;i <= n;i++)
//            printf("%d ",b[i]);
        iniRMQ(n,b);
        while(m--)
        {
            int s,t,ans;
            scanf("%d%d",&s,&t);
            int tl = get_bi(s,t);
//            printf("%d
",tl);
            ans = t- tl+1;
            if(tl-1 > s)
                ans = max(ans,RMQ(s,tl-1));
            printf("%d
",ans);

        }

    }
    return 0;
}