POJ 2452 RMQ + binary

Solution:

Enumerate each position I and find the position j on the Right of I that is smaller than a [I ].

Locate the maximum value between I to j-1 k. If a [k]> a [I], update the answer.

 

 

[Cpp]
# Include <iostream>
# Include <algorithm>
# Include <cstring>
# Include <string>
# Include <cstdio>
# Include <cmath>
# Include <queue>
# Include <map>
# Include <set>
# Define eps 1e-5
# Define maxn55555
# Define MAXM 111111
# Define INF 1000000000
Using namespace std;
Int mi [MAXN] [17], mx [MAXN] [17], w [MAXN];
Int Log [MAXN];
Int n;
Void rmqinit (int n)
{
For (int I = 1; I <= n; I ++) mi [I] [0] = mx [I] [0] = I;
Int m = (int) (log (n * 1.0)/log (2.0 ));
For (int I = 1; I <= m; I ++)
For (int j = 1; j <= n; j ++)
{
Mx [j] [I] = mx [j] [I-1];
Mi [j] [I] = mi [j] [I-1];
If (j + (1 <(I-1) <= n)
{
If (w [mx [j] [I] <w [mx [j + (1 <(I-1)] [I-1]) mx [j] [I] = mx [j + (1 <(I-1)] [I-1];
If (w [mi [j] [I]> w [mi [j + (1 <(I-1)] [I-1]) mi [j] [I] = mi [j + (1 <(I-1)] [I-1];
}
}
}
Int rmqmin (int l, int r)
{
Int m = Log [r-l + 1];
If (w [mi [l] [m]> w [mi [r-(1 <m) + 1] [m]) return mi [r-(1 <m) + 1] [m];
Else return mi [l] [m];
}
Int rmqmax (int l, int r)
{
Int m = Log [r-l + 1];
If (w [mx [l] [m] <w [mx [r-(1 <m) + 1] [m]) return mx [r-(1 <m) + 1] [m];
Else return mx [l] [m];
}
Int bin (int x, int l, int r)
{
Int ret =-1;
While (l <= r)
{
Int m = (l + r)> 1;
If (w [x] <w [rmqmin (l, m)])
L = m + 1, ret = max (ret, m );
Else r = m-1;
}
Return ret;
}
 
Int main ()
{
Log [1] = 0;
For (int I = 2; I <MAXN; I ++) Log [I] = Log [I> 1] + 1;
While (scanf ("% d", & n )! = EOF)
{
For (int I = 1; I <= n; I ++) scanf ("% d", & w [I]);
Rmqinit (n );
Int ans =-1;
For (int I = 1; I <= n; I ++)
{
Int r = bin (I, I + 1, n );
Int k =-1;
If (r> I) k = rmqmax (I, r );
If (w [k]> w [I])
Ans = max (ans, k-I );
}
If (ans <1) ans =-1;
Printf ("% d \ n", ans );
}
Return 0;
}

# Include <iostream>
# Include <algorithm>
# Include <cstring>
# Include <string>
# Include <cstdio>
# Include <cmath>
# Include <queue>
# Include <map>
# Include <set>
# Define eps 1e-5
# Define maxn55555
# Define MAXM 111111
# Define INF 1000000000
Using namespace std;
Int mi [MAXN] [17], mx [MAXN] [17], w [MAXN];
Int Log [MAXN];
Int n;
Void rmqinit (int n)
{
For (int I = 1; I <= n; I ++) mi [I] [0] = mx [I] [0] = I;
Int m = (int) (log (n * 1.0)/log (2.0 ));
For (int I = 1; I <= m; I ++)
For (int j = 1; j <= n; j ++)
{
Mx [j] [I] = mx [j] [I-1];
Mi [j] [I] = mi [j] [I-1];
If (j + (1 <(I-1) <= n)
{
If (w [mx [j] [I] <w [mx [j + (1 <(I-1)] [I-1]) mx [j] [I] = mx [j + (1 <(I-1)] [I-1];
If (w [mi [j] [I]> w [mi [j + (1 <(I-1)] [I-1]) mi [j] [I] = mi [j + (1 <(I-1)] [I-1];
}
}
}
Int rmqmin (int l, int r)
{
Int m = Log [r-l + 1];
If (w [mi [l] [m]> w [mi [r-(1 <m) + 1] [m]) return mi [r-(1 <m) + 1] [m];
Else return mi [l] [m];
}
Int rmqmax (int l, int r)
{
Int m = Log [r-l + 1];
If (w [mx [l] [m] <w [mx [r-(1 <m) + 1] [m]) return mx [r-(1 <m) + 1] [m];
Else return mx [l] [m];
}
Int bin (int x, int l, int r)
{
Int ret =-1;
While (l <= r)
{
Int m = (l + r)> 1;
If (w [x] <w [rmqmin (l, m)])
L = m + 1, ret = max (ret, m );
Else r = m-1;
}
Return ret;
}

Int main ()
{
Log [1] = 0;
For (int I = 2; I <MAXN; I ++) Log [I] = Log [I> 1] + 1;
While (scanf ("% d", & n )! = EOF)
{
For (int I = 1; I <= n; I ++) scanf ("% d", & w [I]);
Rmqinit (n );
Int ans =-1;
For (int I = 1; I <= n; I ++)
{
Int r = bin (I, I + 1, n );
Int k =-1;
If (r> I) k = rmqmax (I, r );
If (w [k]> w [I])
Ans = max (ans, k-I );
}
If (ans <1) ans =-1;
Printf ("% d \ n", ans );
}
Return 0;
}