UVA 11235 Frequent values（RMQ）

Frequent values

TimeLimit:3000Ms

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 Specification

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 Specification

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

Sample Output

题意：给出一个非降序排列的数组A1，A2。A3，……，An，对于一系列询问（i，j），输出Ai，A(i+1)，……，Aj中出现次数最多的值出现的次数。

分析：由于整个数组是非降序的，全部相等元素会聚集在一起，这样就能够把这个数组进行游标编码。比方-1,1,1,2,2,2,4就能够编码成（-1,1），（1,2），（2,3），（4,1），当中（a，b）表示有b个连续的a。

用value[i]和count[i]分别表示第i段的数值和出现次数，num[p]、left[p]、right[p]分别表示位置p所在段的编号和左右端点位置，则每次查询时的结果为下面三部分的最大值：从L到L所在段的结束处的元素个数（即right[L]-L+1）、从R所在段的開始处到R处的元素个数（即R-left[R]+1）、中间第num[L]+1段到第num[R]-1段的count的最大值。这样问题就差点儿转化为了RMQ问题。

#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<iostream>
using namespace std;
const int N = 1e5 + 10;
int n, tot, Q;
int dp[N][20];
int num[N], cnt[N], Left[N], Right[N];
void RMQ_Init()
{
    memset(dp, 0, sizeof(dp));
    for(int i = 1; i <= tot; i++)
        dp[i][0] = cnt[i];
    for(int j = 1; (1<<j) <= n; j++)
        for(int i = 1; i + (1<<j) - 1 <= tot; 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 v, last_v, i;
    while(~scanf("%d",&n))
    {
        if(n == 0) break;
        scanf("%d",&Q);
        tot = 0;
        memset(Left, 0, sizeof(Left));
        memset(Right, 0, sizeof(Right));
        memset(cnt, 0, sizeof(cnt));
        for(i = 1; i <= n; i++)
        {
            scanf("%d",&v);
            if(i == 1)
            {
                ++tot;
                last_v = v;
                Left[tot] = 1;
            }
            if(last_v == v)
            {
                num[i] = tot;
                cnt[tot]++;
                Right[tot]++;
            }
            else
            {
                num[i] = ++tot;
                cnt[tot]++;
                Left[tot] = Right[tot] = i;
                last_v = v;
            }
        }
        RMQ_Init();
        int L, R;
        for(int i = 0; i < Q; i++)
        {
            scanf("%d%d",&L,&R);
            if(num[L] == num[R])
                printf("%d
", R - L + 1);
            else
            {
                int tmp1 = Right[num[L]] - L + 1;
                int tmp2 = R - Left[num[R]] + 1;
                int tmp3 = RMQ(num[L] + 1, num[R] - 1);
                printf("%d
",max(tmp1, max(tmp2, tmp3)));
            }
        }
    }
    return 0;
}