hdoj 5532 Almost Sorted Array

Almost Sorted Array

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1716    Accepted Submission(s): 455

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
这个需要二分求解，不然超时！
   AC代码：
     
#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;#define  N 100100int  a[N],b[N],t[N];int  Find(int left,int right,int val)//二分查找 适合位置 {int  mid;while(right>=left){mid=(left+right)/2;if(t[mid]>val)right=mid-1;elseleft=mid+1;    }    return left;}int main(){int  T;scanf("%d",&T);while(T--){int  n,m,j,i; scanf("%d",&n);for(i=1;i<=n;i++){scanf("%d",&a[i]);b[n+1-i]=a[i];}t[1]=a[1];int  top=1;for(i=2;i<=n;i++){if(a[i]>=t[top])//从小到大 ，，正向 t[++top]=a[i];else{int  temp=Find(1,top+1,a[i]);t[temp]=a[i];}}if(n-top<=1){printf("YES\n");continue;}top=1;t[1]=b[1];for(i=2;i<=n;i++)//从大到小 {if(b[i]>=t[top])t[++top]=b[i];else{int temp=Find(1,top+1,b[i]);t[temp]=b[i];}}if(n-top<=1)    printf("YES\n");    else    printf("NO\n");}return 0;} 


超时代码：
    
#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;#define N  100100#define  INF 0x3f3f3fint main(){    int  T,n,m;    scanf("%d",&T);    while(T--)    {        int  i,j;        int a[N];        scanf("%d",&n);        for(i=1;i<=n;i++)        scanf("%d",&a[i]);        int  dp1[N],dp2[N];        memset(dp1,0,sizeof(dp1));         memset(dp2,0,sizeof(dp2));        for(i=n-1;i>0;i--)          for(j=i+1;j<=n;j++)         {             if(a[j]<=a[i]&&dp1[i]<=dp1[j])//从小到大             dp1[i]=dp1[j]+1;            if(a[j]>=a[i]&&dp2[i]<=dp2[j])//从大到小             dp2[i]=dp2[j]+1;         }         int  max1=0,max2=0;         for(i=1;i<=n;i++)         {             if(max1<dp1[i])             {                 max1=dp1[i];//              count1++;            }             if(max2<dp2[i])             {                  max2=dp2[i];//              count2++;            }         }        if(max1<=1||max2<=1)         printf("YES\n");         else         printf("NO\n");    }    return  0;}