如题，用二分确定左右端点

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define double long double
inline ll read()
{
	ll s=0,w=1;
	char c=getchar();
	while(c<'0'||c>'9'){if(c=='-') w=-1;c=getchar();}
	while(c>='0'&&c<='9'){s=s*10+(c-'0'),c=getchar();}
	return s*w;
}
inline void out(ll x)
{
	ll w=x,c[10005],len=0;
	if(w<0) putchar('-'),w=-w;
	if(w==0){putchar('0');return;}
	while(w) len++,c[len]=w%10+'0',w/=10;
	while(len) putchar(c[len]),len--;
}
const ll M=1e5+5,N=1e9+5;
ll t,n,m,L,V,d[M],v[M],a[M],p[M];
ll cnt,del[M],ans;
struct node
{
	ll l,r;
}b[M];
bool cmp(node x,node y)
{
	return (x.l==y.l?x.r>y.r:x.l<y.l);
}
int main()
{
	t=read();
	while(t--)
	{
		cnt=0;
		n=read(),m=read(),L=read(),V=read();
		for(int i=1;i<=n;i++)
		{
			d[i]=read(),v[i]=read(),a[i]=read();
		}
		for(int i=1;i<=m;i++)
		{
			p[i]=read();
		}
		for(int i=1;i<=n;i++)
		{
			if(v[i]<=V&&a[i]<=0) continue;
			if(a[i]>0)
			{
				ll s;
				if(v[i]>V) s=d[i]-1;
				else s=(V*V-v[i]*v[i])/2/a[i]+d[i];
				ll l=1,r=m,ans=0,mid;
				while(l<=r)
				{
					mid=(l+r)>>1;
					if(p[mid]>s)
					{
						r=mid-1;
						ans=mid;
					}
					else l=mid+1;
				}
				if(ans)
				{
					b[++cnt].l=ans,b[cnt].r=L;
					//cout<<d[i]<<" "<<v[i]<<a[i]<<endl;
					//cout<<ans<<" "<<L<<endl<<endl;
				}
			}
			else
			{
				ll l=1,r=m,ans1=0,mid,ans2=0;
				while(l<=r)
				{
					mid=(l+r)>>1;
					if(p[mid]>=d[i])
					{
						r=mid-1;
						ans1=mid;
					}
					else l=mid+1;
				}
				if(!ans1) continue;
				l=ans1,r=m;
				while(l<=r)
				{
					mid=(l+r)/2;
					double s=sqrt(v[i]*v[i]*1.0+2.0*a[i]*(p[mid]-d[i]));
					if(s>V)
					{
						l=mid+1;
						ans2=mid;
					}
					else r=mid-1;
				}
				if(ans1<=ans2)
				{
					b[++cnt].l=ans1,b[cnt].r=ans2;
					//cout<<d[i]<<" "<<v[i]<<a[i]<<endl;
					//cout<<ans<<" "<<L<<endl<<endl;
				}
			}
		}
		sort(b+1,b+1+cnt,cmp);
		ll pos=N;
		for(int i=cnt;i>=1;i--)
		{
			if(pos<=b[i].r)
			{
				del[i]=1;
			}
			pos=min(pos,b[i].r);
		}
		ans=0,pos=0;
		for(int i=1;i<=cnt;i++)
		{
			if(del[i])
			{
				del[i]=0;
				continue;
			}
			if(pos<b[i].l)
			{
				ans++,pos=b[i].r;
			}
		}
		out(cnt),putchar(' '),out(m-ans),putchar('\n');
	}
	return 0;
}