POJ 3368-Frequent values（RMQ+離散化-最頻繁的元素）

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.

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.

For each query, print one line with one integer: The number of occurrences of the most frequent value within the given range.

#include<iostream>
#include<cstdio>
#include<iomanip>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<map>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define INF 0x3f3f3f3f
#define MAXN 100010
int r[MAXN];//元素個數
int dp[MAXN][20];
int n,q;
struct node
{
    int x;//元素值
    int pos;//離散化之後的元素位置
} a[MAXN];
void rmq()
{
    int temp=(int)(log((double)n)/log(2.0));
    for(int i=0; i<n; i++)
        dp[i][0]=r[i];
    for(int j=1; j<=temp; j++)
        for(int i=0; i<=n-(1<<j); i++)
            dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}

int Maxmum(int s,int e)//計算s~e之間的最大元素
{
    int k=(int)(log((double)e-s+1)/log(2.0));
    return max(dp[s][k],dp[e-(1<<k)+1][k]);
}
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("G:/cbx/read.txt","r",stdin);
    //freopen("G:/cbx/out.txt","w",stdout);
#endif
    /*ios::sync_with_stdio(false);
    cin.tie(0);
    */
    while(cin>>n>>q)
    {
        if(n==0) break;
        memset(a,0,sizeof(a));
        memset(r,0,sizeof(r));
        memset(dp,0,sizeof(dp));
        scanf("%d",&a[0].x);
        r[0]=1;
        int cnt=0;//離散化的位置下標
        for(int i=1; i<n; ++i)//下標0-n-1
        {
            scanf("%d",&a[i].x);
            if(a[i].x==a[i-1].x)
            {
                r[i]=r[i-1]+1;//統計當前位置上的數出現了多少次
                a[i].pos=cnt;//離散化
            }
            else
            {
                r[i]=1;
                ++cnt;//位置下標更新
                a[i].pos=cnt;
            }
        }
        rmq();//dp預處理
        for(int i=0; i<q; ++i)
        {
            int s,e;
            scanf("%d%d",&s,&e);
            --s,--e;
            if(a[s].pos==a[e].pos)  printf("%d\n",e-s+1);//同一區間
            else if(abs(a[s].pos-a[e].pos)==1)//相鄰區間
            {
                int t=s;
                while(t<=e)//找到間隔點
                {
                    if(a[t].pos!=a[t+1].pos) break;
                    ++t;
                }
                printf("%d\n",max((t-s+1),r[e]));
            }
            else if(abs(a[s].pos-a[e].pos)>1)//不相鄰區間
            {
                int t=s;
                while(t<=e)//找到最前面的一個區間的間隔點
                {
                    if(a[t].pos!=a[t+1].pos) break;
                    ++t;
                }
                printf("%d\n",max(t-s+1,Maxmum(t+1,e)));
            }
        }
    }
    return 0;
}