hdu-1806 Frequent values（RMQ，求区间最大频率）

Frequent values

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1475    Accepted Submission(s): 540

Problem Description

You are given a sequence of n integers a ₁ , a ₂ , ... , 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 a ₁ , ... , 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
    
    
    
    
 
    
    
    
    
题意：给一个长度为N的序列，求s~t之间出现最多的数字出现的次数
    
    
    
    
思路：用a数组存题目给的序列，b数组存后面序列有多少个数字和本身一样，再加上1（本身）
    
    
    
    
最后求的时候比一下求最大值就好。用RMQ求区间（b数组上）最大值。
    
    
    
    
代码：
    
    
    
    

    
    
    
    
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 100010
int a[N]; int b[N]; int dp[N][20]; void makermq(int n) { for(int i=0; i<n; i++)
        dp[i][0]=b[i]; for(int j=1; (1<<j)<=n; j++) { for(int i=0; i+(1<<j)-1<n; i++) {
            dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]); } } } int getmax(int s,int t) { int k=(int)(log(t-s+1.0)/log(2.0)); return max(dp[s][k],dp[t-(1<<k)+1][k]); } int finds(int s,int t) { int i=s; while(1) { if(i>=t||a[i]!=a[i+1]) break;
       i++; } return i; } int main() { int n,q; int k,l; while(scanf("%d",&n)&&n) {
        scanf("%d",&q); for(int i=0; i<n; i++)
            scanf("%d",&a[i]); int t=1;
        b[0]=1; for(int i=1; i<n; i++) { if(a[i]==a[i-1])
                t++; else
                t=1;
            b[i]=t; }
        makermq(n); while(q--) {
            scanf("%d %d",&k,&l);
            k--;
            l--; int tm=finds(k,l); int num=tm-k+1; if(tm+1>=l)
                printf("%d\n",num); else
            printf("%d\n",max(num,getmax(tm+1,l))); } } return 0; }