http://acm.split.hdu.edu.cn/showproblem.php?pid=5532

題意：由n個數組成的序列，如果去掉一個數後仍保持非遞增或者非遞減，則輸出YES，否則輸出NO.

思路：只需要求最長非遞增子序列的長度和最長非遞減子序列的長度，如果其中一個長度+1>=n，說明可以實現。(註意：用O(n*n)的算法會導致tle,此處用的是O（nlogn）的算法）

#include<stdio.h>
#define N 100010

int num[N],end1[N],end2[N];

int find1(int flag,int *end,int
 
 low,int high)
{//end1數組按非遞減順序存儲每一位的最小值 
    int mid;
    while(low < high)
    {
        mid = (low+high)>>1;
        if(end[mid] > flag)
            high = mid;
        else
            low = mid + 1;
    }
    return low;
}

int find2(int flag,int *end,int low,int high)
{
 
//end2數組按非遞增順序存儲每一位的最大值 
    int mid;
    while(low < high)
    {
        mid = (low+high)>>1;
        if(end[mid] >= flag)
            low = mid+1;
        else
            high = mid;
    }
    return low;
}

int main()
{
    int t,n,i,j,ans,len1,len2;
    scanf("%d",&t);
    
 
while(t--)
    {
        scanf("%d",&n);
        for(i = 1; i <= n; i ++)
            scanf("%d",&num[i]);
        end1[1] = end2[1] = num[1];
        len1 = len2= 1;
        for(i = 2; i <= n; i ++)
        {
            if(num[i] >= end1[len1])//最長非遞減 
                ans = ++len1;
            else
                ans = find1(num[i],end1,1,len1+1);
            end1[ans] = num[i]; 
            
            if(num[i] <= end2[len2])//最長非遞增 
                ans = ++len2;
            else
                ans = find2(num[i],end2,1,len2+1);
            end2[ans] = num[i];
        }
        if(len1 + 1== n||len2 + 1== n||len1==n||len2==n)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}

hdu 5532【最長非遞增子序列 時間復雜度 nlogn】