POJ2533 LIS模板题

POJ2533

最长上升子序列模板题

Description

A numeric sequence of ai is ordered if a1 < a2 < … < aN. Let the subsequence of the given numeric sequence ( a1, a2, …, aN) be any sequence ( ai1, ai2, …, aiK), where 1 <= i1 < i2 < … < iK <= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).

Your program, when given the numeric sequence, must find the length of its longest ordered subsequence.

Input

The first line of input file contains the length of sequence N. The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated by spaces. 1 <= N <= 1000

Output

Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence.

Sample Input

7
1 7 3 5 9 4 8

Sample Output

4

代码1

复杂度n^2算法
#include
#include
#include
#include
using namespace std;

int a[1005],b[1005];

int main()
{
   int n;
   while (scanf("%d",&n)!=EOF)
   {
       memset(a,0,sizeof(a));
       memset(b,0,sizeof(b));
       for (int i = 0;iscanf("%d",&a[i]);
       b[0] = 1;
       for (int i = 0;i1;
          for (int j = 0;jif (a[i] > a[j] && b[j]+1 > b[i])
                        b[i] = b[j] + 1;
            }
        }
        int maxn = b[0];
        for (int i = 0;iif (b[i] > maxn)
              maxn = b[i];
        printf("%d\n",maxn);
   }
}

代码2

复杂度 nlogn算法
#include
#include
#include
#include
using namespace std;

int find(int *a,int len,int n)
{
    int left = 0,right = len,mid = (left+right)/2;
    while (left <= right)
    {
        if (n > a[mid])
                left = mid+1;
        else if (n < a[mid])
                right = mid - 1;
        else
                return mid;
        mid = (left + right) / 2;
    }
    return left;
}


int main()
{
        int i,j,n,a[100],c[100],len;
        while(scanf("%d",&n))
        {
            for (int i = 0;i"%d",&a[i]);
            c[0] = -1;
            c[1] = a[0];
            len = 1;
            for (int i = 1;ilen,a[i]);
                c[j] = a[i];
                if (j > len)
                        len = j;
            }
            printf("%d\n",len);
        }
}