Frequent values - POJ 3368 RMQ

Frequent values

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 13365		Accepted: 4918

Description

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.

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 , ... , 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：在一个区间里找相同的数的最大个数。

思路：首先离散化把相同的数合并，并且可以得到对于一个数他的左边界和右边界l[i]和r[i]，每次查询特殊处理边界，然后中间的部分用MAX[k][v]表示v及其前2^k的最大值。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int val[100010],l[100010],r[100010],num[100010],point[200030],parent[30][100010],MAX[30][100010];
int solve(int a,int b)
{ if(b<a || b<0 || a<0)
   return 0;
  if(a==b)
   return num[a];
  int len=b-a,k=0;
  while((1<<k)<=len)
   k++;
  k--;
   return max(MAX[k][b],solve(a,parent[k][b]));
}
int main()
{ int n,m,i,j,k,u,v,pos,sum,a,b,c,ans;
  while(~scanf("%d",&n) && n)
  { scanf("%d",&m);
    for(i=1;i<=n;i++)
    { scanf("%d",&val[i]);
      val[i]+=100010;
    }
    val[0]=val[n+1]=0;
    for(i=1;i<=n;i++)
     if(val[i]==val[i-1])
      l[i]=l[i-1];
     else
      l[i]=i;
    for(i=n;i>=1;i--)
     if(val[i]==val[i+1])
      r[i]=r[i+1];
     else
      r[i]=i;
    pos=1;sum=0;
    while(pos<=n)
    { sum++;
      point[val[pos]]=sum;
      num[sum]=r[pos]-l[pos]+1;
      pos=r[pos]+1;
    }
    point[0]=-1;
    for(i=1;i<=sum;i++)
    { parent[0][i]=i-1;
      MAX[0][i]=max(num[i-1],num[i]);
    }
    parent[0][1]=-1;
    for(k=0;k<29;k++)
     for(v=1;v<=sum;v++)
      if(parent[k][v]<0)
      { parent[k+1][v]=-1;
        MAX[k+1][v]=MAX[k][v];
      }
      else
      { parent[k+1][v]=parent[k][parent[k][v]];
        MAX[k+1][v]=max(MAX[k][v],MAX[k][parent[k][v]]);
      }
    while(m--)
    { scanf("%d%d",&a,&b);
      ans=0;
      ans=max(ans,min(r[a],b)-a+1);
      ans=max(ans,b-max(a,l[b])+1);
      ans=max(ans,solve(point[val[r[a]+1]],point[val[l[b]-1]]));
      printf("%d\n",ans);
    }
  }
}