HDU5532 Almost Sorted Array

Almost Sorted Array

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 9982    Accepted Submission(s): 2329

Problem Description

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
NO

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int maxn = 1e5 + 5;
int b[maxn];
int n;
int Search(int num, int low, int high)
{
    int mid;
    while (low <= high)
    {
        mid = (low + high) / 2;
        if (num >= b[mid])
            low = mid + 1;
        else
            high = mid - 1;
    }
    return low;
}
int fin(int *a)
{
    int len, pos;
    b[1] = a[1];
    len = 1;
    for (int i = 2; i <= n; i++)
    {
        if (a[i] >= b[len])
        {
            len = len + 1;
            b[len] = a[i];
        }
        else
        {
            pos = Search(a[i], 1, len);
            b[pos] = a[i];
        }
    }
    if (len >= n - 1)
        return true;
    else
        return false;
}
int t[maxn], tt[maxn];
int main()
{
    int cas;
    scanf_s("%d", &cas);
    while (cas--)
    {
        scanf_s("%d", &n);
        for (int i = 1; i <= n; i++)
            scanf_s("%d", t + i);
        for (int i = 1; i <= n; i++)
            tt[n + 1 - i] = t[i];
        bool flag = fin(t);
        flag |= fin(tt);
        if (flag)
            printf("YES\n");
        else
            printf("NO\n");
    }
}

原文地址：https://www.cnblogs.com/jianqiao123/p/11241115.html