代码
//解法1
#include<bits/stdc++.h>
using namespace std;
#define N 105
int a[N], dpu[N], dpd[N];
int main()
{
int k, n, mxu, mxd;
cin >> k;
while(k--)
{
mxu = mxd = 0;
cin >> n;
for(int i = 1; i <= n; ++i)
cin >> a[i];
for(int i = 1; i <= n; ++i)//求最长上升子序列
{
dpu[i] = 1;
for(int j = 1; j < i; ++j)
if(a[i] > a[j])
dpu[i] = max(dpu[i], dpu[j]+1);
mxu = max(mxu, dpu[i]);
}
for(int i = 1; i <= n; ++i)//求最长下降子序列
{
dpd[i] = 1;
for(int j = 1; j < i; ++j)
if(a[i] < a[j])
dpd[i] = max(dpd[i], dpd[j]+1);
mxd = max(mxd, dpd[i]);
}
cout << max(mxu, mxd) << endl;
}
return 0;
}