rt，大致思路跟 chen_zhe 的 std 一样，但是我二分写法不太一样。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5 + 10;
int T, n, m, ans1, ans2, cnt;
struct Car{
	ll d, v, a;
}c[N];
ll p[N], L, V;
struct node{
	int l, r;
	bool operator<(const node & B) const{
		if(r == B.r) return l < B.l;
		return r < B.r;
	}
}s[N];
int myfid1(int l, int r, int i){ //找第一个合格的 
	int mid;
	while(l < r){
		mid = (l + r) >> 1;
		if(c[i].v * c[i].v + 2ll * c[i].a * (p[mid] - c[i].d) > V * V){
			r = mid;
		}
		else l = mid + 1;
	}
	return r;
}
int myfid2(int l, int r, int i){ //找最后一个不合格的 
	int mid;
	while(l < r){
		mid = (l + r) >> 1;
		if(c[i].v * c[i].v + 2ll * c[i].a * (p[mid] - c[i].d) > V * V){
			l = mid + 1;
		}
		else r = mid;
	}
	return r;
}
int main(){
	//freopen("D:\\OIproblem\\detect\\detect5.in", "r", stdin);
	//freopen("D:\\OIproblem\\detect\\myans.ans", "w", stdout);
	scanf("%d", &T);
	while(T--){
		scanf("%d%d%lld%lld", &n, &m, &L, &V);
		for(int i = 1; i <= n; i++){
			scanf("%lld%lld%lld", &c[i].d, &c[i].v, &c[i].a);
		}
		for(int i = 1; i <= m; i++){
			scanf("%lld", &p[i]);
		}
		ans1 = ans2 = cnt = 0;
		for(int i = 1; i <= n; i++){
			int id = lower_bound(p + 1, p + m + 1, c[i].d) - p;
			if(id == m + 1) continue;
			if(c[i].a >= 0){
				if(c[i].v * c[i].v + 2ll * c[i].a * (p[m] - c[i].d) > V * V){
					++ans1;
					int id2 = myfid1(id, m, i);
					s[++cnt] = (node){id2, m};
				}
			}
			else{
				if(c[i].v * c[i].v + 2ll * c[i].a * (p[id] - c[i].d) > V * V){
					++ans1;
					int id2 = myfid2(id, m, i);
					s[++cnt] = (node){id, id2 - 1};
				}
			}
		}
		sort(s + 1, s + cnt + 1);
		//统计答案 
		int lst = 0;
		for(int i = 1; i <= cnt; i++){
			if(s[i].l > lst){
				++ans2;
				lst = s[i].r;
			}
		}
		printf("%d %d\n", ans1, m - ans2);
	}
	return 0;
}

说句闲话

本人场上对了出 #2 样例的一组数据以外的所有样例，然后气急败坏写了如下注释：

I cannot accept #2

but I can accept #1,3,4,5

Why 9 7?

how about 5 8?

应该不会被禁三吧