# 描述

Emily the entrepreneur has a cool business idea: packaging and selling
snowflakes. She has devised a machine that captures snowflakes as they
fall, and serializes them into a stream of snowflakes that flow, one
by one, into a package. Once the package is full, it is closed and
shipped to be sold. The marketing motto for the company is “bags of
uniqueness.” To live up to the motto, every snowflake in a package
must be different from the others. Unfortunately, this is easier said
than done, because in reality, many of the snowflakes flowing through
the machine are identical. Emily would like to know the size of the
largest possible package of unique snowflakes that can be created. The
machine can start filling the package at any time, but once it starts,
all snowflakes flowing from the machine must go into the package until
the package is completed and sealed. The package can be completed and
sealed before all of the snowflakes have flowed out of the machine.

# Input

The first line of input contains one integer specifying the number of
test cases to follow. Each test case begins with a line containing an
integer n, the number of snowflakes processed by the machine. The
following n lines each contain an integer (in the range 0 to 109 ,
inclusive) uniquely identifying a snowflake. Two snowflakes are
identified by the same integer if and only if they are identical. The
input will contain no more than one million total snowflakes.

# Output

For each test case output a line containing single integer, the
maximum number of unique snowflakes that can be in a package.

# Sample Input

``````1
5
1
2
3
2
1
``````

# Sample Output

``````3
``````

# 代碼

``````#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
typedef long long ll;
const int N=2e5+10;
int a[N];
int main()
{
int t,n;
scanf("%d",&t);
while(t--)
{
set<int>s;
scanf("%d",&n);
for(int i=1; i<=n; i++)
{
scanf("%d",&a[i]);
}
int l=1,r=1,ans=1;
s.insert(a[r]);
r++;
while(r<=n)
{
auto it=s.find(a[r]);
if(it==s.end())
{
s.insert(a[r]);
ans=max(ans,r-l+1);
r++;
}
else
{
while(1)
{
s.erase(a[l]);
l++;
if(s.find(a[r])==s.end())
{
s.insert(a[r]);
r++;
break;
}
}
}
}
printf("%d\n",ans);
}
return 0;
}
``````