F

F - Almost Sorted Array 

 

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,…,ana1,a2,…,an, is it almost sorted?

InputThe first line contains an integer TT indicating the total number of test cases. Each test case starts with an integer nn in one line, then one line with nn integers a1,a2,…,ana1,a2,…,an. 

1≤T≤20001≤T≤2000 
2≤n≤1052≤n≤105 
1≤ai≤1051≤ai≤105 
There are at most 20 test cases with n>1000n>1000.OutputFor 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

//需要用到的算法：最长递增子序列


#include <iostream>
#include <cstdio>
#include <memory.h>
#include <algorithm>

#define INF 0x3f3f3f

using namespace std;

const int MAXN = 1e5 + 7;

int dp[MAXN];
int a[MAXN];
int b[MAXN];

int main() {
    int T;
    cin >> T;
    while(T--) {
        int n;
        scanf("%d", &n);
        memset(dp, INF, sizeof dp);
        for(int i = 0; i < n; i++) {
            scanf("%d", &a[i]);
            b[n - i - 1] = a[i];
        }
        //递增
        //依次遍历，将当前元素插入dp数字中适当的位置
        for(int i = 0; i < n; i++) {
            *upper_bound(dp, dp + n, a[i]) = a[i];
        }
        //求出最长递增子序列的长度
        int cnt1 = lower_bound(dp, dp + n, INF) - dp;
        //递减
        memset(dp, INF, sizeof dp);
        for(int i = 0; i < n; i++) {
            *upper_bound(dp, dp + n, b[i]) = b[i];
        }
        int cnt2 = lower_bound(dp, dp + n, INF) - dp;
        //如果最后递增和递减的长度其中有一个大于或等于(n-1)的话，就输出YES，否则输出NO
        if(cnt1 >= (n - 1) || cnt2 >= (n - 1)) {
            printf("YES
");
        } else {
            printf("NO
");
        }
    }
    return 0;
}