hdu 5532【最长非递增子序列 时间复杂度 nlogn】

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

题意：由n个数组成的序列，如果去掉一个数后仍保持非递增或者非递减，则输出YES，否则输出NO.

思路：只需要求最长非递增子序列的长度和最长非递减子序列的长度，如果其中一个长度+1>=n，说明可以实现。(注意：用O(n*n)的算法会导致tle,此处用的是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数组按非递减顺序存储每一位的最小值 
    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数组按非递增顺序存储每一位的最大值 
    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])//最长非递减 
                ans = ++len1;
            else
                ans = find1(num[i],end1,1,len1+1);
            end1[ans] = num[i]; 
            
            if(num[i] <= end2[len2])//最长非递增 
                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;
}