UV 10534 wavio sequence (longest incrementing sub-sequence DP binary)

Wavio Sequence

Wavio is a sequence of integers. It has some interesting properties.

· Wavio is of odd length I. e.L = 2 * n + 1.

· The first(N + 1)Integers of wavio sequence makes a strictly increasing sequence.

· The Last(N + 1)Integers of wavio sequence makes a strictly decreasing sequence.

· No two adjacent integers are same in a wavio sequence.

For example1, 2, 3, 4, 5, 4, 3, 2, 0Is an wavio sequence of Length9.1, 2, 3, 4, 5, 4, 3, 2, 2Is not a valid wavio sequence. in this problem, you will be given a sequence of integers. you have to find out the length of the longest wavio sequence which is a subsequence of the given sequence. consider, the given sequence:

1 2 3 2 1 2 3 3 3 2 1 5 4 1 2 2 1.

Here the longest wavio sequence is:1 2 3 4 5 4 3 2 1. So, the output will be9.

 

Input

The input file contains less75Test cases. The description of each test case is given below: input is terminated by end of file.

 

Each set starts with a postive integer,N (1 <= n <= 10000). In next few lines there will beNIntegers.

 

Output

For each set of input print the length of longest wavio sequence in a line.

Sample input output for sample input

10 1 2 3 4 5 4 3 2 1 10 19 1 2 3 2 1 2 3 4 3 2 1 5 4 1 2 3 2 2 1 5 1 2 3 4 5

9 9 1

Problemsetter: Md. kamruzzaman, member of elite problemsetters 'panel

The maximum value of the maximum values in the longest ascending sequence (LIS) and longest descending sequence (LDS) of a sequence.

I started to write it directly with DP, And then it timed out. Later I saw someone saying that I would use binary to optimize the time complexity to O (N * logn ).

Use a stack s to save the smallest tail of LIS with the length of I. s [I] top is the top of the stack, that is, the length of the current Lis. Initial s [1] = A [1] Top = 1 Traversal whole sequence when a [I]> S [Top], a [I] goes into the stack in [I] = Top

Otherwise, search for the first subscript POS with a value greater than or equal to a [I] in the stack and replace it, which increases the growth potential of LIS in [I] = Pos;

In [I] indicates the length of lis at the end of a [I;

The process of finding the LDs is similar.

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N = 10005;int a[N], in[N], de[N], s[N], m, n, top, pos;int BinSearch (int k, int le, int ri){    while (le <= ri)    {        m = (le + ri) >> 1;        if (s[m] >= k) ri = m - 1;        else le = m + 1;    }    return ri + 1;}void lis(){    memset (s, 0, sizeof (s));    memset (in, 0, sizeof (in));    s[1] = a[1];    in[1] = top = 1;    for (int i = 2; i <= n; ++i)    {        if (s[top] < a[i])        {            s[++top] = a[i];            in[i] = top;        }        else        {            pos = BinSearch (a[i], 1, top);            s[pos] = a[i];            in[i] = pos;        }    }}void lds(){    memset (s, 0, sizeof (s));    memset (de, 0, sizeof (de));    s[1] = a[n];    de[n] = top = 1;    for (int i = n - 1; i >= 1; --i)    {        if (s[top] < a[i])        {            s[++top] = a[i];            de[i] = top;        }        else        {            pos = BinSearch (a[i], 1, top);            s[pos] = a[i];            de[i] = pos;        }    }}int main(){    while (scanf ("%d", &n) != EOF)    {        for (int i = 1; i <= n; ++i)            scanf ("%d", &a[i]);        int ans = 1;        lis();        lds();        for (int i = 1; i <= n; ++i)        {            if (min (de[i], in[i]) > ans)                ans = min (de[i], in[i]);        }        printf ("%d\n", ans * 2 - 1);    }    return 0;}

And the DP version of TLE.

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N=10005;int a[N],c[N],d[N],n;int dpde(int i){    if(d[i]) return d[i];    d[i]=1;    for(int j=i;j<=n;++j)    {        if(a[i]>a[j])            d[i]=max(d[i],dpde(j)+1);    }    return d[i];}int dpin(int i){    if(c[i]) return c[i];    c[i]=1;    for(int j=i;j>=1;--j)    {        if(a[i]>a[j])            c[i]=max(c[i],dpin(j)+1);    }    return c[i];}int main(){    while(scanf("%d",&n)!=EOF)    {        for(int i=1;i<=n;++i)            scanf("%d",&a[i]);        int ans=1;        memset(d,0,sizeof(d));        memset(c,0,sizeof(c));        for(int i=1;i<=n;++i)        {            if(min(dpde(i),dpin(i))>ans)                ans=min(d[i],c[i]);        }        printf("%d\n",ans*2-1);    }    return 0;}

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UV 10534 wavio sequence (longest incrementing sub-sequence DP binary)