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): 5587    Accepted Submission(s): 1315

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

332 1 733 2 153 1 4 1 5

 

Sample Output

YESYESNO

题意：在原数列中删除一个数判断是否有序；

解：LIS的n*log(n) 啪啪啪打脸

#include<cstdio>#include<cstdlib>#include<cstring>#include<iostream>#include<cmath>#include<string>#include <bits/stdc++.h>using namespace std;typedef long long LL;const int N =2e6+10;const int inf = 0x3f3f3f3f;int a[N], b[N], c[N];int cmp(int A,int B){return A>B;}int main(){    int t, n;    scanf("%d", &t);    while(t--)    {        scanf("%d", &n);        for(int i=0;i<n;i++) scanf("%d", &a[i]), b[n-i-1]=a[i];        int k=1;        c[0]=a[0];        for(int i=1;i<n;i++)        {            if(a[i]>=c[k-1]) c[k++]=a[i];            else  *upper_bound(c,c+k,a[i])=a[i];        }        if(k>=n-1)        {            puts("YES");            continue;        }        k=1;        c[0]=b[0];        for(int i=1;i<n;i++)        {            if(b[i]>=c[k-1]) c[k++]=b[i];            else  *upper_bound(c,c+k,b[i])=b[i];        }        if(k>=n-1)        {            puts("YES");            continue;        }        puts("NO");    }    return 0;}