UVa 10534 Wavio Sequence（LIS N*logN）

题意：

找一个最长（假设长度为2N-1）的子序列,使得前N个元素递增，后N个元素递减。

思路：

LIS。从1-n遍历求出最长上升子序列，再从n-1遍历求出逆序的最长上升子序列。

用最朴素的LIS算法是O(n*n)，这一题就tle了。于是网上搜索出来一种O(nlogn)的算法。

http://www.slyar.com/blog/longest-ordered-subsequence.html

 

大致思路就是：开辟一个堆栈，栈里面的元素是递增的。如果a[i]比栈顶元素要大，就直接压栈。

如果比栈顶元素小，则从底部起找到第一个不比a[i]小的元素替换之。

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <climits>



#define max(a,b) (((a) > (b)) ? (a) : (b))

#define min(a,b) (((a) < (b)) ? (a) : (b))



const int MAXN = 10010;

int stack[MAXN];

int d1[MAXN], d2[MAXN];

int a[MAXN];



int main()

{

    int n;

    while (scanf("%d", &n) != EOF)

    {

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

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

        

        int top = 0;

        stack[0] = INT_MIN;

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

        {

            if (a[i] > stack[top])

                stack[++top] = a[i];

            else

            {

                int low = 1, high = top;

                while (low <= high)

                {

                    int mid = (low + high) >> 1;

                    if (a[i] > stack[mid])

                        low = mid + 1;

                    else

                        high = mid - 1;

                }

                stack[low] = a[i];

            }

            d1[i] = top;

        }



        top = 0;

        stack[0] = INT_MIN;

        for (int i = n; i >= 1; --i)

        {

            if (a[i] > stack[top])

                stack[++top] = a[i];

            else

            {

                int low = 1, high = top;

                while (low <= high)

                {

                    int mid = (low + high) >> 1;

                    if (a[i] > stack[mid])

                        low = mid + 1;

                    else

                        high = mid - 1;

                }

                stack[low] = a[i];

            }

            d2[i] = top;

        }

        int ans = 0;

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

            if (ans < min(d1[i], d2[i]))

                ans = min(d1[i], d2[i]);

        

        printf("%d\n", 2 * ans - 1);

    }

    return 0;

}