Almost Sorted Array　HDU - 5532

最新推荐文章于 2021-07-31 22:01:54 发布

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最新推荐文章于 2021-07-31 22:01:54 发布

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分类专栏： HDU

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本文介绍了一种算法，用于判断一个数组是否为几乎有序数组。几乎有序数组是指可以通过删除一个元素使剩余部分变为有序的数组。文章通过示例输入输出展示了算法的应用，并提供了完整的C++代码实现。

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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

Sample Output

YES
YES
NO

Think：相邻的三个数就可以判断出来是否要删数，把升序和降序两种情况都跑一遍，之后判断一下就行了

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
int a[100010];
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        for (int i=0;i<n;i++){
            scanf("%d",&a[i]);
        }
        int flag=0;
        int sum=0;
        if (a[0]>a[1]){
            flag=1;
        }
        for (int i=2;i<n;i++){
            if (a[i]<a[i-1]){
                if (flag==1){
                    flag=2;
                    break;
                }
                else{
                    if (a[i]>=a[i-2]){
                        flag=1;
                    }
                    else if (i<n&&a[i+1]>=a[i-1]){
                        flag=1;
                        i++;
                    }
                    else if (i==n-1){
                        flag=1;
                    }
                    else{
                        flag=2;
                        break;
                    }
                }
            }
        }
        sum+=flag;
        flag=0;
        if (a[0]<a[1]){
            flag=1;
        }
        for (int i=2;i<n;i++){
            if (a[i]>a[i-1]){
                if (flag==1){
                    flag=2;
                    break;
                }
                else{
                    if (a[i]<=a[i-2]){
                        flag=1;
                    }
                    else if (i<n&&a[i+1]<=a[i-1]){
                        flag=1;
                        i++;
                    }
                    else if (i==n-1){
                        flag=1;
                    }
                    else{
                        flag=2;
                        break;
                    }
                }
            }
        }
        sum+=flag;
        if (sum<4){
            printf("YES\n");
        }
        else{
            printf("NO\n");
        }
    }
    return 0;
}