POJ 3368 （RMQ）

Frequent values

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15924		Accepted: 5794

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

题意：有n个数字，m次查询，每次询问区间[L,R]中出现最多次数的数字，输出最多次数。

题解：先说思路，我们每次查询[L,R]区间，左区间L有可能不是某个数字X的开始位置，那么我们可能会多算了一些长度，那么我们就要设法把这不完整的一段删去，那么我们可以这么处理，把X的[最后出现的下标-当前下标]，这段长度就是X的长度，去掉了数字X的干扰，使得查找区间变为[最后出现X的下标+1，y ]。这里需要考虑一个特殊情况，就是如果区间[L,R]是属于同一个数字，那么就是输出(R-L+1)。详情见下图：

先使用结构体数组记录记录每个数字是第几次出现和数字本身。然后记录每个数字最后出现的下标（建议使用map是最好的），

#include<cstdio>  
#include<cstring>  
#include<cstdlib>  
#include<cmath>  
#include<iostream>  
#include<algorithm>  
#include<vector>  
#include<map>  
#include<set>  
#include<queue>  
#include<string>  
#include<bitset>  
#include<utility>  
#include<functional>  
#include<iomanip>  
#include<sstream>  
#include<ctime>  
using namespace std;

#define N int(1e5+10)  
#define inf int(0x3f3f3f3f)  
#define mod int(1e9+7)  
typedef long long LL;


#ifdef CDZSC  
#define debug(...) fprintf(stderr, __VA_ARGS__)  
#else  
#define debug(...)   
#endif  
int dmax[N][32];

void RMQ(vector< pair<int, int> >&A)
{
	int n = A.size();
	for (int i = 0; i<n; i++)dmax[i][0] = A[i].first;
	for (int j = 1; (1 << j) <= n; j++)
	{
		for (int i = 0; i + (1 << j) - 1<n; i++)
		{
			dmax[i][j] = max(dmax[i][j - 1], dmax[i + (1 << (j - 1))][j - 1]);
		}
	}
}

int query_max(int L, int R)
{
	int k = 0;
	while ((1 << (k + 1)) <= R - L + 1)k++;
	return max(dmax[L][k], dmax[R - (1 << k) + 1][k]);
}

vector<pair<int, int> >v;
int a[N];
map<int, int>w;
int main()
{
#ifdef CDZSC  
	freopen("i.txt", "r", stdin);
	//freopen("o.txt","w",stdout);  
	int _time_jc = clock();
#endif  
	int n, m, x, y;
	while (~scanf("%d", &n))
	{
		w.clear();v.clear();
		if (n == 0)break;
		scanf("%d", &m);
		for (int i = 0; i<n; i++)scanf("%d", &a[i]);
		int last = a[0];
		w[a[0]] = 0;
		v.push_back(make_pair(1, a[0]));
		for (int i = 1; i<n; i++)
		{
			if (last == a[i])
			{
				v.push_back(make_pair(v[i - 1].first + 1, a[i]));
			}
			else   
			{
				v.push_back(make_pair(1, a[i]));
				last = a[i];
			}
			w[a[i]] = i;
		}
		RMQ(v);

		while (m--)
		{
			int ans;
			scanf("%d%d", &x, &y);
			y--; x--;
			if (v[x].second == v[y].second)
			{
				printf("%d\n",y-x+1);
				continue;
			}
			ans = w[v[x].second] - x+1;
			x = w[v[x].second] + 1;
			printf("%d\n", max(ans, query_max(x, y)));
		}
	}
#ifdef CDZSC  
	debug("time: %d\n", int(clock() - _time_jc));
#endif  
	return 0;

}