UVA-11235-Frequent values(RMQ)

UVA-11235
Frequent values
Time Limit: 3000 MS
 
メモリLimit: アンロックnown
 
64 bit IO Format: %lld&%ll
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Description
2007/2008 ACM International Collegate Programming Cotest 
University of Ulm Local Conteest
Problem F:Frequent values
You are given a sequence of n integers a 1 , a 2 , ... , an in non-decrease 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ラインcontains n integers a 1 , ... , an (-100000≦ai ≦100000、for each i∈｛1,…n｝separated by spaces.You can asume that for each i∈｛1,…,n-1｝:ai ≦ai＋1.The follwing q LINE 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 followwed by 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

A naive algorithm may not run in time!
ソurce
Root:Copetive Programming:Increase the Lower Bound of Programming Conttes(Steven&Felix Halim):Chapter 2.Data Structures and Libries:Data Structures With Our-Own Libries:  Segment Tree
Root:Copetive Programming 3:The New Lower Bound of Programming Conttes(Steven&Felix Halim):Data Structures and Libries::Data Structures with Our-Own Libries:  Tree-related Data Structures
Root:AOAPC I:Beginning Algorithm Conttes--Training Gide(Rujia Liu):Chapter 3.Data Structures::Maintaining Interval Data:  Examples
Root:Copetive Programming 2:This increas the lower bound of Programming Conttes.Again(Steven&Felix Halim):Data Structures and Libries:Data Structures with Our-Own Libries:  Tree-related Data Structures
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考え方:長い間やってきましたが、結構です.最後に見つけたのは配列の下付きです.最初から、私のこのコードは合わせて使います.降順ではないので、旅程コード(RLE)を使って、区間内の最大値を算出したらいいです.
ACコード:

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 100005;
int a[maxn];
int value[maxn], coun[maxn];
int num[maxn], le[maxn], ri[maxn];
int d[maxn][400];
int maxnum;//    

void RMQ_init() {
	for(int i = 0; i < maxnum+1; i++) d[i][0] = coun[i];
	for(int j = 1; (1 << j) <= maxnum+1; j++)
		for(int i = 0; i + (1<<j) - 1 < maxnum+1; i++)
			d[i][j] = max(d[i][j-1], d[i + (1 << (j-1))][j-1]);
}

int RMQ(int l, int r) {
	int k = 0;
	while((1<<(k+1)) <= r - l + 1) k++;
	return max(d[l][k], d[r-(1<<k) +1][k]);
}

int main() {
	int n, q;
	while(scanf("%d", &n) && n!=0) {
		scanf("%d", &q);
		for(int i = 0; i < n; i++) 
			scanf("%d", &a[i]);
		
		maxnum = 0;
		int cnt = 1;//       
		value[0] = a[0], coun[0] = cnt, num[0] = 0, le[0] = 0, ri[0] = 0;
		for(int i = 1; i < n; i++) {
			if(a[i] != a[i-1]) {
				ri[maxnum] = i - 1;
				cnt = 1; maxnum++;
				le[maxnum] = i; ri[maxnum] = i;
				num[i] = maxnum;
				value[maxnum] = a[i];
				coun[maxnum] = cnt;
			} 
			else {
				cnt++;
				coun[maxnum] = cnt;
				num[i] = maxnum;
				ri[maxnum] = i;
			}
		} 
		
		RMQ_init();
			
		for(int i = 0; i < q; i++) {
			int l, r;
			scanf("%d %d", &l, &r);
			l--; r--;
			//printf("%d %d %d
", ri[num[l]]-l+1, r-le[num[r]]+1, RMQ(num[l]+1, num[r]-1));
			if(num[l] == num[r]) printf("%d
", r - l + 1);
			else if(num[r] - num[l] == 1) printf("%d
", max(ri[num[l]]-l+1, r-le[num[r]]+1));
			else printf("%d
", max(max(ri[num[l]]-l+1, r-le[num[r]]+1), RMQ(num[l]+1, num[r]-1)));
		}
	}
	return 0;
}