POJ 3368 Frequent values (RMQ)

 

 

Frequent values

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 11540		Accepted: 4206

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

Source

Ulm Local 2007

 

 

 

 

 

 

 

这题需要一个区间内，出现次数最多的数出现的次数。

 

先离散化处理下，然后转化成RMQ问题。

 

 

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;

const int MAXN = 100010;

int dp[MAXN][20];
int mm[MAXN];

void initRMQ(int n,int b[])
{
    mm[0] = -1;
    for(int i = 1;i <= n;i++)
    {
        mm[i] = ((i&(i-1)) == 0)?mm[i-1]+1:mm[i-1];
        dp[i][0] = b[i];
    }
    for(int j = 1;j <= mm[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 = mm[y-x+1];
    return max(dp[x][k],dp[y-(1<<k)+1][k]);
}

int a[MAXN];
int hash[MAXN];
int L[MAXN],R[MAXN];
int b[MAXN];

int main()
{
    int n,m;
    while(scanf("%d",&n) == 1 && n)
    {
        scanf("%d",&m);
        for(int i = 1;i <= n;i++)
            scanf("%d",&a[i]);
        int k = 1;
        int l = 1;
        memset(b,0,sizeof(b));
        for(int i = 1;i <= n;i++)
        {
            if(i > 1 && a[i] != a[i-1])
            {
                for(int j = l;j < i;j++)
                {
                    L[j] = l;
                    R[j] = i-1;
                }
                b[k] = i - l;
                l = i;
                k ++;
            }
            hash[i] = k;
        }
        for(int j = l;j <= n;j++)
        {
            L[j] = l;
            R[j] = n;
        }
        b[k] = n-l+1;
        initRMQ(k,b);

        int u,v;
        while(m--)
        {
            scanf("%d%d",&u,&v);
            int t1 = hash[u];
            int t2 = hash[v];
            if(t1 == t2)
            {
                printf("%d\n",v-u+1);
                continue;
            }
            int ans = max(R[u]-u+1,v-L[v]+1);
            t1++;
            t2--;
            if(t1 <= t2)ans = max(ans,rmq(t1,t2));
            printf("%d\n",ans);
        }
    }
    return 0;
}