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RMQ [HDU 1806] Frequent values

2024-07-28 20:26:44   225人阅读

Frequent values

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1146    Accepted Submission(s): 415


Problem Description
You are given a sequence of n integers a1 , a2 , ... , an 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 ai , ... , aj . 

 

 

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 , ... , an(-100000 ≤ ai ≤ 100000, for each i ∈ {1, ..., n}) separated by spaces. You can assume that for each i ∈ {1, ..., n-1}: ai ≤ ai+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 10 102 31 105 100
 

 

Sample Output
143
Hint
A naive algorithm may not run in time!
 

 

Source
HDOJ 2007 Summer Exercise（1）
 
分成三段、中间RMQ、然后求最大值即可
 
#include <iostream>#include <cstdio>#include <cmath>using namespace std;#define N 100010int n,m;int a[N];int dp[N][20];int id[N];int len[N],l[N],r[N];void init(){    int i,j;    for(i=1;i<=n;i++)    {        dp[i][0]=len[i];    }    int k=(int)(log((double)n)/log(2.0));    for(j=1;j<=k;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 query(int i,int j){    int k=(int)(log((double)(j-i+1))/log(2.0));    int res=max(dp[i][k],dp[j-(1<<k)+1][k]);    return res;}int main(){    int i,pos;    while(scanf("%d",&n),n)    {        pos=0;        scanf("%d",&m);        for(i=1;i<=n;i++)        {            scanf("%d",&a[i]);            if(a[i]!=a[i-1])            {                pos++;                l[pos]=i;            }            id[i]=pos;            r[pos]=i;            len[pos]=r[pos]-l[pos]+1;        }        n=pos;        init();        int x,y,xx,yy,ans1,ans2,ans3;        while(m--)        {            int x,y;            scanf("%d%d",&x,&y);            xx=id[x];            yy=id[y];            if(xx==yy) ans1=ans2=y-x+1; //特殊情况、当x和y在同一个区间、答案是y-x+1            else            {                ans1=r[xx]-x+1;                ans2=y-l[yy]+1;            }            ans3=0;            if(xx+1<=yy-1)ans3=query(xx+1,yy-1);            printf("%d\n",max(max(ans1,ans2),ans3));        }    }    return 0;}

 

RMQ [HDU 1806] Frequent values


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