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;
}