Almost Sorted Array

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2504    Accepted Submission(s): 616

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）

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题意：依次给你n个数字，判断其是否是至少>=n-1长的单调子序列；

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
typedef  long long  ll;
typedef unsigned long long ull;
#define MM(a,b) memset(a,b,sizeof(a));
#define inf 0x7f7f7f7f
#define FOR(i,n) for(int i=1;i<=n;i++)
#define CT continue;
#define PF printf
#define SC scanf
const int mod=1000000007;
const int N=100000+10;
ull seed=13331;

int dp[N],ans[N],a[N];
int main()
{
    int cas;
    scanf("%d",&cas);
    while(cas--){
       int n;scanf("%d",&n);
       for(int i=1;i<=n;i++) scanf("%d",&a[i]);
       int len=1,flag1=0,flag2=0;
       ans[1]=a[1];
       for(int i=2;i<=n;i++)
       {
             if(a[i]>=ans[len]) {len++;ans[len]=a[i];}
             else{
                int pos=upper_bound(ans+1,ans+len,a[i])-ans;
                ans[pos]=a[i];
             }
       }
       if(len>=n-1) flag1=1;
       len=1;ans[1]=a[n];
       for(int i=n-1;i>=1;i--)
       {
             if(a[i]>=ans[len]) {len++;ans[len]=a[i];}
             else{
                int pos=upper_bound(ans+1,ans+len,a[i])-ans;
                ans[pos]=a[i];
             }
       }
       if(len>=n-1) flag2=1;
       if(flag1||flag2) printf("YES\n");
       else printf("NO\n");
    }
    return 0;
}

　　分析：严格单调递增子序列就是lower_bound（），直接进行替换；非严格单调子序列就是upper_bound();

 if(a[i]>=ans[len]) {len++;ans[len]=a[i];}
             else{
                int pos=upper_bound(ans+1,ans+len,a[i])-ans;//非严格单调
                ans[pos]=a[i];
             }