POJ 2533

最长上升子序列裸题
在网上看到有两种方法...一种复杂度O(N^2)，一种O(NlogN)。orz

O(N^2)：

#include<cstdio>

#define N 1001

int main()

{

    int n,ans;

    int a[N],d[N];



    while(~scanf("%d",&n)){

    for(int i=1; i<=n; i++)

        scanf("%d",&a[i]);



    ans=0;

    for(int i=1; i<=n; i++)

    {

        d[i]=1;

        for(int j=1; j<=i-1; j++)

        {

            if(a[i]>a[j] && d[i]<d[j]+1)

                d[i]=d[j]+1;

        }

        if(d[i]>ans)

            ans=d[i];

    }

    printf("%d\n",ans);

    }

    return 0;

}

View Code

O(NlogN)：

 #include <cstdio>  

    using namespace std;  

      

    const int maxn = 1000 + 10;  

    int n, num[maxn], LIS[maxn];  

      

    int binary_search(int start, int end, int target)  

    {  

        while (end > start) {  

            int mid = (end + start) / 2;  

            if (target > LIS[mid])  

                start = mid + 1;  

            else  

                end = mid;  

        }  

        return start;  

    }  

      

    int lis()  

    {  

        LIS[1] = num[1];  

        int maxl = 1;  

        for (int i = 2; i <= n; ++i) {  

            if (num[i] > LIS[maxl])  

                LIS[++maxl] = num[i];  

            else {  

                int pos = binary_search(1, maxl, num[i]);  

                LIS[pos] = num[i];  

            }  

        }  

        return maxl;  

    }  

      

    int main()  

    {  

        scanf("%d", &n);  

        for (int i = 1; i <= n; ++i)  

            scanf("%d", &num[i]);  

        printf("%d\n", lis());  

    }  

View Code