通过树状数组维护时候,记得添加初始元素0,使得维护元素为空的情况(也就是代表前缀和是s[0])
#include<bits/stdc++.h>
#define ll long long
#define endl '\n'
#define PII pair<int,int>
using namespace std;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
const double eps=1e-8;
const int N=5e5+5;
int n,cnt,dp[N],c[N];
ll a[N],pre[N];
set<ll> st;
map<ll,int> id;
void add(int x,int k)
{
while(x<=n)
c[x]=max(c[x],k),x+=(x&-x);
}
int query(int x)
{
int cnt=-inf;
while(x>0)
cnt=max(cnt,c[x]),x-=(x&-x);
return cnt;
}
int main()
{
std::ios::sync_with_stdio(false);
int t;
cin >> t;
while(t--)
{
cin >> n;
cnt = 0;
st.clear();
id.clear();
for(int i = 1 ; i <= n;i++)
{
c[i] = -inf;
dp[i] = -inf;
}
st.insert(0);
for(int i = 1 ; i <= n ;i++)
{
cin >> a[i];
pre[i] = pre[i - 1] + a[i];
st.insert(pre[i]);
}
for(ll p : st)
{
id[p] = ++cnt;
}
add(id[0],0); //代表空元素
for(int i = 1 ; i <= n ; i++)
{
if(a[i] == 0) dp[i] = dp[i - 1];
else if(a[i] < 0) dp[i] = dp[i - 1] - 1;
dp[i] = max(dp[i],query(id[pre[i]] - 1) + i);
add(id[pre[i]],dp[i] - i);
}
cout << dp[n] << endl;
}
return 0;
}