捞，由于改用差分约束处理点覆盖时答案正确（只是超时），所以初步认为问题出在这步上。但好像还是没有从中发现问题 /kk

#include <bits/stdc++.h>
using namespace std;
int T, n, m, L, V, d[100001], v[100001], a[100001], p[100001];
struct interval {
    int l, r;
    bool operator<(const interval &x) const {
        return l < x.l;
    }
} t[100001];
int findl(int i, int l, int r) {
    while (l <= r) {
        int mid = (l + r) >> 1;
        if ((p[mid] - d[i]) * a[i] * 2 > V * V - v[i] * v[i]) r = mid - 1;
        else l = mid + 1;
    }
    return l;
}
int findr(int i, int l, int r) {
    while (l <= r) {
        int mid = (l + r) >> 1;
        if ((p[mid] - d[i]) * a[i] * 2 <= V * V - v[i] * v[i]) r = mid - 1;
        else l = mid + 1;
    }
    return r;
}
int main() {
    #ifndef ONLINE_JUDGE
        freopen("detect.in", "r", stdin);
        freopen("detect.out", "w", stdout);
    #endif
    scanf("%d", &T);
    while (T--) {
        scanf("%d%d%d%d", &n, &m, &L, &V);
        for (int i = 1; i <= n; i++) scanf("%d%d%d", &d[i], &v[i], &a[i]);
        for (int i = 1; i <= m; i++) scanf("%d", &p[i]);
        int cnt = 0, ans1 = 0, ans2 = 1;
        for (int i = 1; i <= n; i++) {
            int pos = lower_bound(p + 1, p + m + 1, d[i]) - p;
            if (pos > m) continue;
            if (v[i] > V && a[i] >= 0) {
                t[++cnt] = (interval){pos, m};
                ans1++;
            }
            else if (v[i] <= V && a[i] > 0) {
                int l = findl(i, pos, m);
                if (l <= m) {
                    t[++cnt] = (interval){l, m};
                    ans1++;
                }
            }
            else if (v[i] > V && a[i] < 0) {
                int r = findr(i, pos, m);
                if (pos <= r) {
                    t[++cnt] = (interval){pos, r};
                    ans1++;
                }
            }
        }
        sort(t + 1, t + cnt + 1);
        for (int i = 2, R = t[1].r; i <= cnt; i++) {
            if (t[i].l > R) {
                R = t[i].r;
                ans2++;
            }
            else R = min(R, t[i].r);
        }
        printf("%d %d\n", ans1, m - ans2);
    }
    return 0;
}