最长子是先增加再减少 [英] Longest subsequence that first increases then decreases
问题描述
我试图解决以下问题:
一个序列中的元素首先减少的值,然后增加被称为V-序列。在一个有效的V-序列应该有向上升臂下降和至少一个元素的至少一种元素。
例如,5 3 1 9 17 23是具有在减少臂即5和3中增加臂即9,17及23的两个元素,和3个元素的有效的V-序列。但没有序列6 4 2或8 10 15的垂直序列,因为6 4 2有,而8 10 15已在减少部分没有元素的增加部分不元素。
的序列的子序列中,删除从序列零个或多个元件获得。例如定义7,2 10,8 2 7 6,8 2 7 10 6等等是8 2 7 10 6
的有效的子序列给定N数序列发现其最长的子序列,其是V型序列
目前,我有一个为O(n ^ 2)的解决方案,其中我第一次初始化数组(M []),使得每个m [I]包含最长递增序列开始'我'在阵列中。
类似地,我初始化另一个阵列(D []),使得各d [I]包含最长递减序列ENDING在这一点上。
这两种方法需要为O(n ^ 2)
我现在通过这些阵列,并选择m的最大值[I] + D [I] -1,使得所需要的条件得到满足。
我想知道的是 - 是否有一个O(N LG N)解决方案?因为我的解决方案不要求在时间限制运行。谢谢:)
code:
#包括< cstdio>
#包括<算法>
使用名字空间std;
INT M [200000]。
INT D [200000]。
INT N;
INT ARR [200000]。
无效LIS()
{
米[N-1] = 1;
INT maxvPos = -1;
INT MAXV = -1;
的for(int i = N-2; I> = 0;我 - )
{
MAXV = -1;
为(诠释J = i + 1的; J&n种; J ++)
{
如果((M [J] +1> MAXV)及及(ARR [1] - ,常用3 [J]))
{
MAXV = M [J] +1;
maxvPos = j的;
}
}
如果(MAXV大于0)
{
M [] = MAXV;
}
其他
米[I] = 1;
}
}
无效LDS()
{
D [0] = 1;
INT MAXV = -1;
INT maxvPos = -1;
的for(int i = 1;我n种;我++)
{
MAXV = -1;
对于(INT J = I-1; J> = 0; j--)
{
如果((D [J] +1> MAXV)及和放大器;常用3 [J] GT;常用3 [I])
{
MAXV = D [J] +1;
maxvPos = j的;
}
}
如果(MAXV大于0)
D [i] = MAXV;
其他
D [I] = 1;
}
}
INT解决()
{
LIS();
LDS();
INT MAXV = 0;
INT CURR = 0;
的for(int i = 0;我n种;我++)
{
CURR = D [I] + M [] -1;
如果((四[I] 0)&安培;及(米[I]&0))
{
如果(CURR!= 1)
MAXV = MAX(CURR,MAXV);
}
}
返回MAXV;
}
/ *静态无效printArr(INT []一)
{
对于(INT I:1)
System.out.print第(i +);
的System.out.println();
} * /
诠释的main()
{
scanf函数(%d个,和放大器; N);
的for(int i = 0;我n种;我++)
{
scanf函数(%d个,和放大器;常用3 [I]);
}
的printf(%D \ N,解决了());
返回0;
}
有 O(NlgK)
算法最长递增序列问题,其中氏/ code>是LIS的长度。您可以检查维基百科,在算法的描述。 LightOJ 也有一个很好的教程(这可能需要登录)。
I am trying to solve the following question :
A sequence in which the value of elements first decrease and then increase is called V- Sequence. In a valid V-Sequence there should be at least one element in the decreasing and at least one element in the increasing arm.
For example, "5 3 1 9 17 23" is a valid V-Sequence having two elements in the decreasing arm namely 5 and 3, and 3 elements in the increasing arm namely 9, 17 and 23 . But none of the sequence "6 4 2" or "8 10 15" are V-Sequence since "6 4 2" has no element in the increasing part while "8 10 15" has no element in the decreasing part.
A sub-sequence of a sequence is obtained by deleting zero or more elements from the sequence. For example definition "7", "2 10", "8 2 7 6", "8 2 7 10 6" etc are valid sub-sequences of "8 2 7 10 6"
Given a sequence of N numbers find its longest sub-sequence which is a V-Sequence.
I currently have an O( n^2 ) solution wherein I first initialize an array ( m[] ) such that each m[i] contains the longest increasing sequences STARTING at 'i' within the array.
Similarly, I initialize another array ( d[] ), such that each d[i] contains the longest decreasing sequence ENDING at that point.
Both of these operations take O( n^2 )
I now go through these arrays and choose the maximum value of m[i] + d[i] -1 , such that required conditions are satisfied.
What I want to know is - Is there an O( n lg n ) solution ?? Because my solution does not run within required time limits. Thank you :)
CODE :
#include<cstdio>
#include<algorithm>
using namespace std;
int m[ 200000 ];
int d[200000 ];
int n;
int arr[200000 ];
void LIS()
{
m[ n-1 ] = 1;
int maxvPos = -1;
int maxv = -1;
for( int i=n-2; i>=0; i-- )
{
maxv = -1;
for( int j=i+1; j<n; j++ )
{
if( ( m[j]+1 > maxv ) && ( arr[i] < arr[j]) )
{
maxv = m[j]+1;
maxvPos = j;
}
}
if( maxv>0 )
{
m[i] = maxv;
}
else
m[i ] = 1;
}
}
void LDS()
{
d[0] = 1;
int maxv = -1;
int maxvPos = -1;
for( int i=1; i<n; i++ )
{
maxv = -1;
for( int j=i-1; j>=0; j-- )
{
if( ( d[j]+1 > maxv) && arr[j]>arr[i] )
{
maxv = d[j]+1;
maxvPos = j;
}
}
if( maxv>0 )
d[i] = maxv;
else
d[i]=1;
}
}
int solve()
{
LIS();
LDS();
int maxv = 0;
int curr = 0;
for( int i=0; i<n; i++ )
{
curr = d[i] + m[i] -1 ;
if( ( d[i]>0) && (m[i]>0 ))
{
if( curr != 1 )
maxv = max( curr, maxv );
}
}
return maxv;
}
/* static void printArr( int[] a )
{
for( int i : a )
System.out.print( i + " ");
System.out.println();
} */
int main()
{
scanf( "%d", &n );
for( int i=0; i<n; i++ )
{
scanf("%d", &arr[i] );
}
printf("%d\n", solve() );
return 0;
}
There are O(NlgK)
algorithm for Longest Increasing Subsequence problem, where K
is the LIS length. You can check Wikipedia for a description of the algorithm. LightOJ also has a nice tutorial (this might require login).
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