HDU 5532 Almost Sorted Array (dp)

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探讨如何判断一个数组是否可以通过移除一个元素变为非严格递增或递减的序列。介绍了解题思路及实现方法。

Almost Sorted Array

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2508 Accepted Submission(s): 619

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

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：这题题意应该比较清晰，就是问你给出队列是否可以取去一个数后变成非严格递增或非严格递减的序列，可以输出YES，不行输出NO。

解题思路：先假可以，那么该子序列的非严格单调上升长度为len1，非严格单调递减子序列长度为len2，满足max(len1,len2)>=n，所以只要求出这两个长度即可，然后加以判断。

<span style="font-size:18px;">#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
const int maxn=100005;
int a[maxn],b[maxn];
vector<int> v;
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        v.clear();
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        vector<int>::iterator it=v.begin();
        for(int i=0;i<n;i++)
        {
            if((it=upper_bound(v.begin(),v.end(),a[i]))==v.end())
                 v.push_back(a[i]);
            else
                *it=a[i];
        }
        int ans=v.size();
        v.clear();
        for(int i=n-1;i>=0;i--)
        {
            if((it=upper_bound(v.begin(),v.end(),a[i]))==v.end())
                v.push_back(a[i]);
            else
                *it=a[i];
        }
        int ans1=v.size();
        if(ans>=n-1 || ans1>=n-1)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}</span>