1102. Sticks problem

 

Xuanxuan has n sticks of different length. One day, she puts all her sticks in a line, represented by s1, s2, s3, ..., sn. After measuring the length of each stick sk (1 ≤ k ≤ n), she finds that for some sticks si and sj (1 ≤ i < j ≤ n), each stick placed between si and sj is longer than si but shorter than sj.

Now given the length of s1, s2, s3, ..., sn, you are required to find the maximum value j – i.

 

Input. Contains multiple test cases. Each case contains two lines. First line is a single integer n (n ≤ 50000), indicating the number of sticks. Second line contains n different positive integers (not larger than 105), indicating the length of each stick in order.

 

Output. Output the maximum value j – i in a single line. If there is no such i and j, just output -1.

 

Sample input

Sample output

4

5 4 3 6

4

6 5 4 3

9

12 4 8 7 5 9 6 3 1

1

-1

4

 

 

 

SOLUTION

RMQ + binary search

 

Algorithm analysis

For each index of i, find the maximum index k (i ≤ k ≤ n) for which RMinQ(si+1, …, sk) > si with binary search. The desired j is the index at which RMaxQ(si, …, sk) (i ≤ j ≤ k) is achieved.

Thus, for each index of i, find the maximum possible index j and among all such pairs (i, j) calculate the maximum difference j – i.

 

Example

Consider the third sample.

Let i = 2 (s2 = 4). The largest index is k = 7, for which RMinQ(si+1, …, sk) > 4. RMaxQ(s3, …, s7) is achieved at index j = 6.

 

Algorithm realization

 

#include <cstdio>

#include <algorithm>

#define MAX 50010

#define LOGMAX 16

using namespace std;

 

int dp_max[MAX][LOGMAX], dp_min[MAX][LOGMAX];

int a[MAX];

int i, j, n, res;

 

void Build_RMQ_Array(int *b)

{

  int i, j;

  for (i = 1; i <= n; i++)

  {

    dp_min[i][0] = b[i];

    dp_max[i][0] = i;

  }

 

  for (j = 1; 1 << j <= n; j++)

    for (i = 1; i + (1 << j) - 1 <= n; i++)

    {

      if (b[dp_max[i][j - 1]] > b[dp_max[i + (1 << (j - 1))][j - 1]])

        dp_max[i][j] = dp_max[i][j - 1];

      else

        dp_max[i][j] = dp_max[i + (1 << (j - 1))][j - 1];

 

      dp_min[i][j] =

        min(dp_min[i][j - 1], dp_min[i + (1 << (j - 1))][j - 1]);

    }

}

 

int RangeMaxQuery(int i, int j)

{

  int k = 0;

  while ((1 << (k + 1)) <= j - i + 1) k++;

 

  if (a[dp_max[i][k]] > a[dp_max[j - (1<<k) + 1][k]])

    return dp_max[i][k];

  else

    return dp_max[j - (1<<k) + 1][k];

}

 

int RangeMinQuery(int i, int j)

{

  int k = 0;

  while ((1 << (k + 1)) <= j - i + 1) k++;

  return min(dp_min[i][k],dp_min[j - (1<<k) + 1][k]);

}

 

int BinSearch(int Left, int Right)

{

  int MinValue = a[Left++];

  if (RangeMinQuery(Left,Right) > MinValue) return Right;

 

  while (Right > Left)

  {

    int Middle = (Left + Right) / 2;

    if (RangeMinQuery(Left,Middle) > MinValue)

      Left = Middle + 1;

    else

      Right = Middle;

  }

 

  if (dp_min[Left][0] <= MinValue) Left--;

  return Left;

}

 

int main(void)

{

  while(scanf("%d",&n) == 1)

  {

    for(i = 1; i <= n; i++) scanf("%d",&a[i]);

 

    Build_RMQ_Array(a);

    res = 0;

 

    for(i = 1; i <= n; i++)

    {

      j = BinSearch(i, n);

      j = RangeMaxQuery(i,j);

      if (j - i > res) res = j - i;

    }

    if (res == 0) res = -1;

    printf("%d\n",res);

  }

  return 0;

}