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 , ... , 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

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

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
[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
;
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;

}