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
0
Sample Output
1
4
3
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\n", 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\n",max(tmp1, max(tmp2, tmp3)));
}
}
}
return 0;
}
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