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  • 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;
    }

     

    -------------------------------------------------------

    kedebug

    Department of Computer Science and Engineering,

    Shanghai Jiao Tong University

    E-mail: kedebug0@gmail.com

    GitHub: http://github.com/kedebug

    -------------------------------------------------------

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  • 原文地址:https://www.cnblogs.com/kedebug/p/2774113.html
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