HDU 5532 Almost Sorted Array（最长不下降子序列nlogn模板）

布满荆棘的人生 2022-06-08 09:55 187阅读 0赞

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 a1,a2,…,an, is it almost sorted?

Input

The first line contains an integer T T indicating the total number of test cases. Each test case starts with an integer n n in one line, then one line with n n integers a1,a2,…,an a1,a2,…,an.  
  
1≤T≤2000 1≤T≤2000  
2≤n≤105 2≤n≤105  
1≤ai≤105 1≤ai≤105  
There are at most 20 test cases with n>1000 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

题解：

题意：

给你一个序列，让你从里面去掉一个使获得的序列为不上升或者为不下降序列

思路：

直接求最长不下降和不上升序列，看长度值是否为n或者n-1，模板走起

代码：

#include<iostream>
    #include<cstring>
    #include<stdio.h>
    #include<math.h>
    #include<string>
    #include<stdio.h>
    #include<queue>
    #include<stack>
    #include<map>
    #include<vector>
    #include<deque>
    #include<algorithm>
    #define ll long long
    #define INF 100861111
    #define M (t[k].l+t[k].r)/2
    #define lson k*2
    #define rson k*2+1
    using namespace std;
    const int N = 100001;
    int  a[N],b[N],f[N], d[N];	// f[i]用于记录 a[0...i]的最大长度
    int bsearch(const int *f, int size, const int &a)
    {
        int  l=0, r=size-1;
    	while( l <= r )
    	{
            int  mid = (l+r)>>1;
    		if( a >= d[mid-1] && a < d[mid] )
    			return mid;				// >&&<= 换为: >= && <
    		else if( a <d[mid] )
    				r = mid-1;
            else l = mid+1;
        }
    }
    int LIS(const int *a, const int &n)
    {
    	int  i, j, size = 1;
    	d[0] = a[0]; f[0] = 1;
    	for( i=1; i < n; ++i )
    	{
    		if( a[i] < d[0] )			 // <= 换为: <
    			j = 0;
    		else if( a[i] >= d[size-1] ) // > 换为: >=
    			j = size++;
    		else
    			j = bsearch(d, size, a[i]);
    
    		d[j] = a[i]; f[i] = j+1;
    	}
    	return size;
    }
    int main()
    {
        int n,i,test;
        scanf("%d",&test);
        while(test--)
        {
            scanf("%d",&n);
            for(i=0;i<n;i++)
            {
                scanf("%d",&a[i]);
                b[n-i-1]=a[i];
            }
            if(LIS(a,n)>=n-1||LIS(b,n)>=n-1)
            {
                printf("YES\n");
                continue;
            }
            printf("NO\n");
        }
        return 0;
    }