【POJ 3368】Frequent values

Frequent values

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15880		Accepted: 5778

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

题目大意：求区间内重复次数最多的元素所重复的次数。

思路：由于数据的范围很大，所以是一道很典型的RMQ，由于数组保证非降序，所以相同元素一定排列在一起，这样我们就可以将相同元素划分在一个区间内，用一个num数组储存第i个元素在新划分后的第一个小区间，left数组储存第i个小区间的左端点，right数组储存右端点，res存第i个区间有多少个元素，这样在求解时如果发现输入的l和r值在一个区间内，答案即为r-l+1，若不在一个区间内，答案即为right[num[l]]-l+1,r-left[num[r]]+1,RMQ(num[l]+1,num[r]-1))三者中的最大值。

代码：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#define N 100010
using namespace std;
int num[N],lef[N],rig[N],res[N],dp[N][20],cnt,n;
int maxn(int a,int b,int c)
{
    return max(max(a,b),c);
}
void RMQ_init()
{
    memset(dp,0,sizeof(dp));
    for(int i=1; i<=cnt; i++) dp[i][0]=res[i];
    for(int j=1; (1<<j)<=cnt; j++)
        for(int i=1; i+(1<<j-1)<=cnt; i++)
            dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}
int RMQ(int L,int R)
{
    if(L>R)
        return 0;
    int k=0;
    while((1<<(k+1))<=R-L+1) k++;
    return max(dp[L][k],dp[R-(1<<k)+1][k]);
}
int main()
{
    int q,temp,before;
    while(~scanf("%d",&n)&&n)
    {
        scanf("%d",&q);
        cnt=0;
        memset(num,0,sizeof(num));
        memset(lef,0,sizeof(lef));
        memset(rig,0,sizeof(rig));
        memset(res,0,sizeof(res));
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&temp);
            if(i==1)
            {
                ++cnt;
                lef[cnt]=1;
                before=temp;
            }
            if(temp==before)
            {
                num[i]=cnt;
                res[cnt]++;
                rig[cnt]++;
            }
            else
            {
                num[i]=++cnt;
                res[cnt]++;
                lef[cnt]=rig[cnt]=i;
                before=temp;
            }
        }
        RMQ_init();
        while(q--)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            if(num[l]==num[r])
            {
                printf("%d
",r-l+1);
                continue;
            }
            printf("%d
",maxn(rig[num[l]]-l+1,r-lef[num[r]]+1,RMQ(num[l]+1,num[r]-1)));
        }
    }
}