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 , ... , 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

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
; 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\n",tl);
ans = t- tl+1;
if(tl-1 > s)
ans = max(ans,RMQ(s,tl-1));
printf("%d\n",ans);

}

}
return 0;
}