POJ 3368 Frequent values (基础RMQ)

Frequent values

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14742		Accepted: 5354

Description

You are given a sequence of n integersa₁ , a₂ , ... , a_n in non-decreasing order. In addition to that, you are given several queries consisting of indicesi and j (1 ≤ i ≤ j ≤ n). For each query, determine the most frequent value among the integersa_i , ... , a_j.

Input

The input consists of several test cases. Each test case starts with a line containing two integersn and q (1 ≤ n, q ≤ 100000). The next line containsn integers a₁ , ... , a_n (-100000 ≤ a_i ≤ 100000, for eachi ∈ {1, ..., n}) separated by spaces. You can assume that for each i ∈ {1, ..., n-1}: a_i ≤ a_i+1. The followingq 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 single0.

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

题目链接：http://poj.org/problem?

id=3368

题目大意：有一串数字，查询区间中频数最大的数字的频数

题目分析：由于数字是按非递增序排列好的，我们能够先预处理出某连续数字在当前位置时出现的频数，比方例子有
dp[1]=1，val[1] = -1
dp[2]=2，val[2] = -1
dp[3]=1，val[3] = 1
dp[4]=2，val[4] = 1
dp[5]=3，val[5] = 1
dp[6]=4。val[6] = 1
。。。
则对于查询区间(l，r)。答案即为区间（l。tmp）和（tmp。r）某一数字出现的频数的较大的那个(l <= tmp <= r)
对于区间(l，tmp)直接可得出答案。对于区间(tmp, r)我们能够用RMQ求解

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
int const MAX = 100005;
int st[MAX][20], dp[MAX], val[MAX];
int n, q;

void RMQ_Init()
{
    for(int i = 1; i <= n; i++)
        st[i][0] = dp[i];
    int k = log((double)(n + 1)) / log(2.0);
    for(int j = 1; j <= k; j++)
        for(int i = 1; i + (1 << j) - 1 <= n; i++)
            st[i][j] = max(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
}

int Query(int l, int r)
{
    if(l > r)
        return 0;
    int k = log((double)(r - l + 1)) / log(2.0);
    return max(st[l][k], st[r - (1 << k) + 1][k]);
}

int main()
{
    while(scanf("%d", &n) != EOF && n)
    {   
        scanf("%d", &q);
        dp[1] = 1;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &val[i]);
            if(i > 1)
                dp[i] = (val[i] == val[i - 1] ?
 dp[i - 1] + 1 : 1);
        }
        RMQ_Init();
        while(q--)
        {
            int l, r;
            scanf("%d %d", &l, &r);
            int tmp = l;
            while(tmp <= r && val[tmp] == val[tmp - 1])
                tmp ++;
            printf("%d
", max(Query(tmp, r), tmp - l));
        }
    }
}