HDOJ 5532-Almost Sorted Array【模拟】



Almost Sorted Array

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1443    Accepted Submission(s): 410

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
   
   
   
   

 

Source

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

 

解题思路

       题目大意就是给出一段数字，让你找出如果将这个数组中的一个数抽出，该数组还是不是一个递增或递减的数组。

       可以判断数组d递增递减两个方面，一次遍历这个数组看看是不是能找到两条沟或两条以上，但是同时还要判断这条沟只能是由一个数字构成的4 5 6 1 2 7 8 9这样的由两个数字组成的沟输出就是NO。

#include<stdio.h>
#define INF 0x3f3f3f3f
int map[100005];
bool F1(int x)
{
	int sum=0;
	map[0]=-INF,map[x+1]=INF;
	for(int i=2;i<=x;i++)
	{
		if(map[i]<map[i-1])
		{
			if(sum==1)
			{
				return false;
			}
			sum++;
			if(map[i+1]>=map[i-1]||map[i]>=map[i-2])
			continue;
			else
			return false;
		}
	}
	return true;
}
bool F2(int x)
{
	int sum=0;
	map[0]=INF,map[x+1]=-INF;
	for(int i=2;i<=x;i++)
	{
		if(map[i]>map[i-1])
		{
			if(sum==1)
			return false;
			sum++;
			if(map[i+1]<=map[i-1]||map[i]<=map[i-2])
			continue;
			else
			return false;
		}
	}
	return true;
}
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		int n;
		scanf("%d",&n);
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&map[i]);
		}
		if(F1(n)||F2(n))
		{
			printf("YES\n");
		}
		else
		{
			printf("NO\n");
		}
	}
	return 0;
}