hdu 5532（最长上升子序列）

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

3
3
2 1 7
3
3 2 1
5
3 1 4 1 5

 

Sample Output

YES
YES
NO

题意：给你一组数，能否删除一个后使他成为升序或者降序

思路：

正反分别求一次最长上升子序列，只要长度有一次≥n-1即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <vector>
#include <algorithm>
#include <functional>
typedef long long ll;
using namespace std;
const int maxn = 1e5 + 5;
int b[maxn];
int n;
int Search(int num,int low,int high)
{
    int mid;
    while(low <= high)
    {
        mid = (low+high)/2;
        if(num >= b[mid])
            low= mid+1;
        else
            high = mid-1;
    }
    return low;
}

int fin(int *a)
{
    int len,pos;
    b[1] = a[1];
    len = 1;
    for(int i = 2;i <= n;i++)
    {
        if(a[i] >= b[len])
        {
            len = len+1;
            b[len] = a[i];
        }
        else
        {
            pos = Search(a[i],1,len);
            b[pos ]  = a[i];
        }
    }
    if(len >= n-1)
        return true;
    else
        return false;
}

int t[maxn],tt[maxn];

int main()
{
    int cas;
    scanf("%d",&cas);
    while(cas--)
    {
        scanf("%d",&n);
        for(int i = 1;i <= n;i++)
            scanf("%d",t+i);
        for(int i = 1;i <= n;i++)
            tt[n+1-i] = t[i];
        bool flag = fin(t);
        flag |= fin(tt);
        if(flag)
            printf("YES\n");
        else
            printf("NO\n");
    }
}