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


[align=left]Problem Description[/align]
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 .

 

[align=left]Input[/align]
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. 

 

[align=left]Output[/align]
For each query, print one line with one integer: The number of occurrences of the most frequent value within the given range.

 

[align=left]Sample Input[/align]

10 3
-1 -1 1 1 1 1 3 10 10 10
2 3
1 10
5 10
0

 

[align=left]Sample Output[/align]

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
;
int b
;
int dp
[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;
}