Almost Sorted Array(o(nlgn)求解LIS)

Almost Sorted Array

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 10562    Accepted Submission(s): 2449

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）

 

Recommend

hujie

onlgn求解LIS的基本思想是，用dp[i]保存长度为i的最长子序列的最大值的最小值。
遍历数组，如果a >= dp[len]，接到后面，否则，在dp中寻找第一个大于这a的数，把他替换掉，原因很好想，要想使得序列足够长，那么dp[i]越小越好。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn =  + ;

int n, val[maxn];
int cnt1, cnt2, dp[maxn];

int main() {
   int t;
   int a, temp;
   scanf("%d", &t);
   while(t --) {
      cnt1 = , cnt2 = ;
      memset(dp, , sizeof dp);
      scanf("%d", &n);
      for(int i = ; i < n; i ++) scanf("%d", &val[i]);
      for(int i = ; i < n; i ++) {
         if(val[i] >= dp[cnt1]) dp[++ cnt1] = val[i];
         else {
            temp = upper_bound(dp + , dp +  + cnt1, val[i]) - dp;
            dp[temp] = val[i];
         }
      }
      memset(dp, , sizeof dp);
      for(int i = n - ; i >= ; i --) {
         if(val[i] >= dp[cnt2]) dp[++ cnt2] = val[i];
         else {
            temp = upper_bound(dp + , dp +  + cnt2, val[i]) - dp;
            dp[temp] = val[i];
         }
      }
      if(cnt1 >= n -  || cnt2 >= n - ) printf("YES\n");
      else printf("NO\n");
   }
   return ;
}