For the daily milking, Farmer John's N cows (1 ≤
N ≤ 50,000) always line up in the same order. One day Farmer John decides
to organize a game of Ultimate Frisbee with some of the cows. To keep things
simple, he will take a contiguous range of cows from the milking lineup to play
the game. However, for all the cows to have fun they should not differ too much
in height.
Farmer John has made a list of Q (1 ≤ Q ≤
200,000) potential groups of cows and their heights (1 ≤ height ≤
1,000,000). For each group, he wants your help to determine the difference in
height between the shortest and the tallest cow in the group.
Input. Line 1: Two space-separated integers, N and Q.
Lines 2..N+1: Line i+1
contains a single integer that is the height of cow i
Lines N+2..N+Q+1: Two
integers A and B (1 ≤ A ≤ B ≤ N), representing the range of
cows from A to B inclusive.
Output. Lines 1..Q: Each line contains a single integer that
is a response to a reply and indicates the difference in height between the tallest
and shortest cow in the range.
Sample Input
6 3
1
7
3
4
2
5
1 5
4 6
2 2
Sample
Output
6
3
0
структуры данных – RMQ
В задаче необходимо реализовать Range Minimum / Maximum Query. Каждый запрос –
это разница между максимумом и минимумом на отрезке.
Реализация алгоритма
#include <stdio.h>
#define MAX
50010
#define
LOGMAX 16
int
dp_max[MAX][LOGMAX], dp_min[MAX][LOGMAX];
int
a[MAX];
int i, n,
q, u, v;
int max(int i, int j)
{
return (i
> j) ? i : j;
}
int min(int i, int j)
{
return (i
< j) ? i : j;
}
void
Build_RMQ_Array(int *b)
{
int i, j;
for (i = 1; i
<= n; i++) dp_max[i][0] = dp_min[i][0] = b[i];
for (j = 1; 1
<< j <= n; j++)
for (i = 1;
i + (1 << j) - 1 <= n; i++)
{
dp_max[i][j] =
max(dp_max[i][j - 1], 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 temp, k =
0;
if (i > j)
temp = i, i = j, j = temp;
while ((1
<< (k + 1)) <= j - i + 1) k++;
int _max =
max(dp_max[i][k],dp_max[j - (1<<k) + 1][k]);
int _min =
min(dp_min[i][k],dp_min[j - (1<<k) + 1][k]);
return _max -
_min;
}
int main(void)
{
scanf("%d
%d",&n,&q);
for(i = 1; i
<= n; i++) scanf("%d",&a[i]);
Build_RMQ_Array(a);
for(i = 0; i
< q; i++)
{
scanf("%d
%d",&u,&v);
printf("%d\n",RangeMaxQuery(u,v));
}
return 0;
}