HDU-5532 Almost Sorted Array（LIS）

Almost Sorted Array

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 3883    Accepted Submission(s): 984

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）

题意：给你一段序列，让你判断其是否是 almost sorted ，如果是的话就输出YES，else NO

思路：看看其LIS长度是不是大于等于n-1，要二分，不然会TLE
ps：原来我以前upper_bound写不对是因为里面的地址放错了，sad

/*
题意
*/
#include
#include
#include
#include
#include
using namespace std;
const int N = 100005;
int a[N],dp[N];
int main()
{
    int t,n;
    cin >> t;
    while(t--)
    {
        scanf("%d",&n);
        for(int i = 1;i <= n;i++)
            scanf("%d",&a[i]);//,dp[i] = 1;
        int num = 1,pos;dp[1] = a[1];
        for(int i = 2;i <= n;i++)
            if(a[i] >= dp[num])
                dp[++num] = a[i];
            else
            {
                pos = upper_bound(dp+1,dp+num,a[i])-dp;
                dp[pos] = a[i];
            }

        int len = 1;dp[1] = a[n];
        for(int i = n-1;i >= 1;i--)
            if(a[i] >= dp[len])
                dp[++len] = a[i];
            else
            {
                pos = upper_bound(dp+1,dp+len,a[i])-dp;
                dp[pos] = a[i];
            }

        if(num >= n-1 || len >= n-1)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}