翻了题解区都发现没有我的做法。

大概就是说我当时想第二问可以贪心地去做，枚举监控，考虑是否有一辆车只经过这一个监控，如果有就不删，否则贪心删除。

通过大样例+对拍+官方数据，但是不会证明/wul

类似的考场代码：

#include <bits/stdc++.h>
#define ll long long
#define N 1000010
using namespace std;
ll T, n, m, L, V; 
struct car {
    ll d, v, a, l, r;
} a[N];
ll t[N], p[N];
void upd(ll x, ll v) {
    x ++;
    for(; x <= L + 2; x += x & -x) t[x] += v;
}
ll qry(ll x) {
    x ++;
    ll res = 0;
    for(; x > 0; x -= x & -x) res += t[x];
    return res;
}
void solve() {
    scanf("%lld %lld %lld %lld", &n, &m, &L, &V);
    for(ll i = 1; i <= n; i ++) {
        scanf("%lld %lld %lld", &a[i].d, &a[i].v, &a[i].a);
        if(a[i].a > 0) {
            if(a[i].v * a[i].v + 2 * a[i].a * (L - a[i].d) > V * V) {
                a[i].r = L;
                ll l = a[i].d, r = L;
                while(l <= r) {
                    ll mid = (l + r) >> 1;
                    if(a[i].v * a[i].v + 2 * a[i].a * (mid - a[i].d) > V * V) {
                        a[i].l = mid;
                        r = mid - 1;
                    } else {
                        l = mid + 1;
                    }
                }
            } else {
                a[i].l = a[i].r = L + 1;
            }
        } else if(a[i].a == 0) {
            if(a[i].v > V) {
                a[i].l = a[i].d;
                a[i].r = L;
            } else {
                a[i].l = a[i].r = L + 1;
            }
        } else {
            if(a[i].v > V) {
                a[i].l = a[i].d;
                ll l = a[i].d, r = L;
                while(l <= r) {
                    ll mid = (l + r) >> 1;
                    if(a[i].v * a[i].v + 2 * a[i].a * (mid - a[i].d) > V * V) {
                        a[i].r = mid;
                        l = mid + 1;
                    } else {
                        r = mid - 1;
                    }
                }
            } else {
                a[i].l = a[i].r = L + 1;
            }
        }
    }
    ll ans1 = 0, ans2 = 0;
    for(ll i = 1; i <= L + 2; i ++) t[i] = 0;
    for(ll i = 1; i <= m; i ++) {
        scanf("%lld", &p[i]);
        upd(p[i], 1);
    }
    for(ll i = 1; i <= n; i ++) {
        if(qry(a[i].r) - qry(a[i].l - 1)) {
            ans1 ++;
        }
    }
    sort(a + 1, a + n + 1, [](car a, car b) {
        return a.l < b.l;
    });
    for(ll i = 1; i <= L + 2; i ++) t[i] = 0;
    ll pos1 = 1, pos2 = 1;
    p[m + 1] = L + 1;
    for(ll i = 1; i <= m; i ++) {
        while(pos1 <= n && a[pos1].l <= p[i]) {
            upd(a[pos1].r, 1);
            pos1 ++;
        }
        if(qry(p[i + 1] - 1) - qry(p[i] - 1)) {
            while(pos2 <= n && a[pos2].l <= p[i]) {
                upd(a[pos2].r, -1);
                pos2 ++;
            }
        } else {
            ans2 ++;
        }
    }
    printf("%lld %lld\n", ans1, ans2);
}
int main() {
    scanf("%lld", &T);
    while(T --) {
        solve();
    }
}