POJ 3368 解題報告 RMQ

Frequent values

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10935		Accepted: 4006

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

題目大意：

給出一個連續不降的整數序列，序列長度爲n (n<=1e6)，接着給出q次詢問，每次詢問序列中從l到r中出現次數最多的數的個數。

解題思路：

利用RMQ我的如下，O(N)統計出每個數最右邊和它相等的數的位置，記做rt[i]。

如樣例中的      1 1 -1 -1 -1 -1 3 10 10 10

統計的rt數組是2 1 6   6  6   6 7  10 10 10

然後再計算出每個數到其最左邊和它相等的數的長度

樣例所得的數組c[i]爲  1 2 1 2 3 4 1 1 2 3。

然後對於每次詢問l r，答案就應該是l到其最右邊的數的長度，和l最右邊的下一個數到r中的c[i]的最大值。

比如對於詢問2 4

a[l]的值爲1，到其最右邊的是i=2，長度爲1，然後rt[l]+1=3到r=4中c最大的是2。所以答案就是2。

其中從rt[l]+1到r中求c[i]最大值可以用RMQ（也可以用線段樹）來求。

代碼供參考：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int a[100100];
int rt[100100];
int dp[200100][30];
int log2(int x){
	int res=0;
	while(x){x/=2;res++;}
	return res;
}
int main(){
	int n,q;
	while(~scanf("%d",&n)){
		memset(dp,0,sizeof(dp));
		memset(rt,0,sizeof(rt));
		if(n==0)return 0;
		scanf("%d",&q);
		for(int i=1;i<=n;i++)scanf("%d",&a[i]);
		int now=n;
		rt[n]=n;
		for(int i=n-1;i>=1;i--){
			if(a[i]<a[now])now=i;
			rt[i]=now;
		}
		dp[1][0]=1;
		for(int i=2;i<=n;i++){
			if(a[i]==a[i-1])dp[i][0]=dp[i-1][0]+1;
			else dp[i][0]=1;
		}
		int log2n=log2(n)-1;
		for(int i=1;i<=log2n;i++){
			for(int j=1;j<=n;j++){
				dp[j][i]=max(dp[j][i-1],dp[j+(1<<(i-1))][i-1]);
			}
		}
		while(q--){
			int l,r;
			scanf("%d%d",&l,&r);
			if(rt[l]>=r)printf("%d\n",r-l+1);
			else{
				int tmp=rt[l]-l+1;
				int len=r-rt[l];
				int log2len=log2(len)-1;
				tmp=max(tmp,max(dp[rt[l]+1][log2len],dp[r-(1<<(log2len))+1][log2len]));
				printf("%d\n",tmp);
			}
		}
	}
}