HDU 5532 Almost Sorted Array（LIS）

最新推荐文章于 2019-01-25 21:56:45 发布

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#LIS #数据结构 #编程

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本文介绍了一种高效算法，用于判断一个整数数组是否可以通过移除一个元素变为有序。通过计算最长递增子序列的长度来实现，适用于比赛或实际问题解决。

We are all familiar with sorting algorithms: quick sort, merge sort, heap sort, insertion sort, selection sort, bubble sort, etc. But sometimes it is an overkill to use these algorithms for an almost sorted array.

We say an array is sorted if its elements are in non-decreasing order or non-increasing order. We say an array is almost sorted if we can remove exactly one element from it, and the remaining array is sorted. Now you are given an array a1,a2,…,an
, is it almost sorted?
Input
The first line contains an integer T indicating the total number of test cases. Each test case starts with an integer n in one line, then one line with n integers a1,a2,…,an.

1≤T≤2000
2≤n≤105
1≤ai≤105
There are at most 20 test cases with n>1000
.
Output
For each test case, please output "`YES`" if it is almost sorted. Otherwise, output "`NO`" (both without quotes).
Sample Input

3
3
2 1 7
3
3 2 1
5
3 1 4 1 5

Sample Output

YES
YES

今天组队训练赛遇到的一个题，用一般的DP最长子序列方法作超时到怀疑人生，新学到一种做法，效率更高！

#include <bits/stdc++.h>
#define N 200100
using namespace std;
int a[100100];//顺序储存
int b[100100];//逆序储存
int LIS(int a[], int s);//寻找最长上升子序列
int main()
{
    int n;
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &a[i]);
            b[n - i - 1] = a[i];
        }
        //int x = LIS(a, n);
        //int y = LIS(b, n);
        //printf("%d %d\n", x, y);
        if(LIS(a, n) >= n - 1 || LIS(b, n) >= n - 1)？//如果满足题意则满足最长上升子序列长度 >= n-1
        {
            printf("YES\n");
        }
        else
        {
            printf("NO\n");
        }
    }
    return 0;
}
int LIS(int a[], int n)
{
    int Stack[N];//手动建栈
    int top = -1;
    int index;
    Stack[++top] = a[0];//首位元素入栈
    for(int i = 1; i < n; i++)
    {
        if(a[i] >= Stack[top])满足a[i] >= 栈顶元素则入栈
        {
            Stack[++top] = a[i];
        }
        else
        {
            index = upper_bound(Stack, Stack + top + 1, a[i]) - Stack;//如果不满足条件从栈中寻找第一个大于a[i]的元素并将其替换
            Stack[index] = a[i];
        }
    }
    return top + 1;//栈中元素个数即为最长上升子序列长度
}