poj 3368 Frequent values (RMQ或线段树)

Frequent values

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 16959		Accepted: 6125

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 integersa₁ , ... , 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 qlines 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

求一段数字区间出现次数最多的数字的频率，数列单调不减。对数列转化后可以用RMQ或者线段树求解。

RMQ经常用来查询区间最值问题，对于任意区间，RMQ算法可以方便的在O(1)时间查询出最值。

以最大值为例，考虑递推方程：f[i][j]=max(f[i][j-1],f[i+(1<<(j-1))][j-1]);

f[i][j]:表示区间[i,i+2^j-1],j=0,1,2,在区间内即可。

上面式子即表示：大区间[i,i+2^j-1]区间=小区间[i,i+2^(j-1)-1]U[i+2^(j-1),i+2^(j-1)+2^(j-1)-1]，

等式两边明显相等（相同），故成立。

对于该题，查询区间[l,r]，对于区间左端点处和区间外面相等的数字单独处理即可。

RMQ解法：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<math.h>
#include<string.h>
using namespace std;
#define LL long long int
#define rep(i,a,n) for(int i=a;i<n;++i)
#define per(i,a,n) for(int i=n-1;i>=a;--i
#define N 100005
const int inf=1e9+10;
int a[N],num[N],rt[N];
int dp[N][20];
void rmq(int n)
{
    int i,j;
    rep(i,1,n+1) dp[i][0]=num[i];
    for(j=1;j<20;++j){
        for(i=1;i+(1<<(j-1))<=n;++i){
            dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
        }
    }
}
int myquery(int l,int r)
{
    if(l>r)
        return 0;
    int k,t1,t2;
    k=log(r-l+1.0)/log(2.0);
    t1=dp[l][k];
    t2=dp[r-(1<<k)+1][k];
    return max(t1,t2);
}
int main()
{
    //freopen("in.txt","r",stdin);
    int n,q;
    while(scanf("%d",&n),n){
        scanf("%d",&q);
        a[0]=inf;
        rep(i,1,n+1){
            scanf("%d",&a[i]);
            if(i==1){
                num[i]=1;
            }
            else if(a[i]==a[i-1]){
                num[i]=num[i-1]+1;
            }
            else{
                num[i]=1;
            }
        }
        per(i,1,n+1){
            if(a[i]==a[i+1]){
                rt[i]=rt[i+1];
            }
            else{
                rt[i]=i;
            }
        }
        rmq(n);
        int l,r,t,tmp,ans;
        while(q--){
            scanf("%d%d",&l,&r);
            if(rt[l-1]==rt[l]) t=min(r,rt[l])+1;
            else t=l;
            tmp=myquery(t,r);
            ans=max(t-l,tmp);
            cout<<ans<<endl;
        }
    }
    return 0;
}