HDU 5532 Almost Sorted Array （2015ACM/ICPC长春LIS）

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【题目链接】：click here~~

【题目描述】：

Almost Sorted Array

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 272 Accepted Submission(s): 132

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

【题意】：删除任意一个元素使得数列单调。

【思路】我们考虑删除一个元素后，使得单调递增最长的序列长度满足>=n-1就可以了，单调递减的就翻转数列在操作一遍即可，复杂度(O(nlogN))

代码：

/*
* Problem:HDU 5532
* Running time: 1100MS
* Complier: G++
* Author: javaherongwei
* Create Time: 21:20 2015/11/02 星期一
*/
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
const int inf=1e9+7;
const int maxn=1e5+10;
using namespace std;
inline int max(int a,int b){return a>b?a:b;
}
int a[maxn],b[maxn],c[maxn];
int n;
int binary_search(int x){ //二分搜索int l=1,r=n;while(l<r){int mid=(l+r)>>1;if(b[mid]>x) r=mid;else l=mid+1;}return l;
}
bool LIS(const int *a){ //LISmemset(b,inf,sizeof(b));b[0]=0;int maxx=0;for(int i=1; i<=n; ++i){int idx=binary_search(a[i]);//i位置的LIS为idxmaxx=max(idx,maxx);b[idx]=a[i];}if(maxx>=n-1) return true; //满足条件即可return false;
}
int main(){//freopen("1.txt","r",stdin);int t;scanf("%d",&t);while(t--){scanf("%d",&n);for(int i=1; i<=n; ++i){scanf("%d",&a[i]);c[n-i+1]=a[i];}if(LIS(a)||LIS(c)) puts("YES");else puts("NO");}return 0;
}

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