poj 3368 Frequent values(RMQ)

Frequent values

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17812		Accepted: 6417

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 10
2 3
1 10
5 10
0

Sample Output

1
4
3

//  细节！！你说因为细节浪费了多少时间！！

题意就是给你一堆连续的数，给你区间，让你查询给定区间的数出现频率最大的频率

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define maxn 100005
using namespace std;
int d[maxn][50];
int left[maxn],right[maxn],val[maxn],cnt[maxn],num[maxn];/*left记录第i段的左端点，right记录第i段的又断点*/
int n,m,len;//val记录第i段的值，num记录第i个数是属于第几段
void RMQ()
{
int i,j;
for( i = 1; i <= len; i++)
{
d[i][0]=cnt[i];
}
for(int j = 1; j != 50; ++j)
{
for(int i = 1; i <= n; ++i)
{
if(i + (1 << j) - 1 <= n)
{
d[i][j] = max(d[i][j-1],d[i + (1<<(j-1)) ][j-1]);

}

}

}
}
int get_RMQ(int a,int b)
{
int k = (int)(log(b - a + 1.0) / log(2.0));   //！！括号
int maxsum = max(d[a][k], d[b - (1 << k) + 1][k]);
return maxsum;
}
int main()
{
int i,j;
int a[maxn];
while(~scanf("%d",&n),n)
{
scanf("%d",&m);
memset(cnt,0,sizeof(cnt));
len=1;
scanf("%d",&a[1]);
val[len]=a[1];
cnt[len]=1;
left[1]=1;
num[len]=1;
for(i=2;i<=n;i++)
{
scanf("%d",&a[i]);
if(a[i] == a[i-1])
{
cnt[len]++;
num[i]=len;
}
else
{
right[len] = i - 1;
len++;
left[len] = i;
val[len] = a[i];
cnt[len]=1;
num[i]=len;
}
}
RMQ();
while(m--)
{
int l,r;
int sum=0;
scanf("%d%d",&l,&r);
if(num[l] == num[r])
printf("%d\n",r-l+1);
else
{
int sum = 0;
if(num[r] - num[l] >= 2)//如果存在超过三段的话  就先找中间的 最大的频率
{
sum = get_RMQ(num[l]+1,num[r]-1);
}
sum = max(sum,max( right[ num[l]] - l + 1, r - left[num[r]] + 1 ));//看中间最大的频率与两侧的谁大
printf("%d\n",sum);
}

}
}
return 0;
}