关于S组T2

#include<bits/stdc++.h>
#define MAXN 100010
using namespace std;
struct sec{ double l, r; };
bool operator < (sec a, sec b){ return a.l < b.l; }
bool operator > (sec a, sec b){ return a.l > b.l; }
int t, n, m, len, lim, ans1, ans2;
int d[MAXN], v[MAXN], a[MAXN];
double p[MAXN];
bool check(double l, double r){
    int p1 = lower_bound(p + 1, p + m + 1, l) - p;
    int p2 = upper_bound(p + 1, p + m + 1, r) - p - 1;
    if(p1 > m || p2 < 1) return false;
    if(p1 > p2) return false;
    return true;
}
int main(){
    scanf("%d",&t);
    while(t--){
        ans1 = 0; ans2 = 0;
        vector<sec> s;
        scanf("%d%d%d%d",&n,&m,&len,&lim);
        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("%lf",&p[i]);
        sort(p + 1, p + m + 1);
        for(int i = 1; i <= n; i++){
            if(a[i] > 0){
                int mv = v[i] * v[i] + 2 * a[i] * (len - d[i]);
                if(mv <= lim * lim) continue;
                else{
                    double dis = (lim * lim - v[i] * v[i]) / (a[i] * 2.0);
                    double l = d[i] + dis, r = lim;
                    if(check(l, r)){
                        s.push_back((sec){l, r});
                        ans1++;
                    }
                }
            }else if(a[i] < 0){
                if(v[i] <= lim) continue;
                else{
                    double dis = (lim * lim - v[i] * v[i]) / (a[i] * 2.0);
                    double l = d[i], r = d[i] + dis;
                    if(check(l, r)){
                        s.push_back((sec){l, r});
                        ans1++;
                    }
                }
            }else if(a[i] == 0 && v[i] > lim){
                s.push_back((sec){0, (double)len});
                ans1++;
            }
        }
        sort(s.begin(), s.end());
        for(int i = 0; i < s.size(); ){
            double nl = s[i].l, nr = s[i].r;
            i++;
            while(s[i].l < nr && check(max(nl, s[i].l), min(nr, s[i].r))){
                nl = max(nl, s[i].l);
                nr = min(nr, s[i].r);
                i++;
            }
            ans2++;
        }
        printf("%d %d\n",ans1, m - ans2);
    }
    return 0;
}
/*
3
5 5 15 3
0 3 0
12 4 0
1 1 4
5 5 -2
6 4 -4
2 5 8 9 15
5 5 15 3
0 3 0
12 4 0
1 1 4
5 5 -2
6 4 -4
2 5 8 9 15
5 5 15 3
0 3 0
12 4 0
1 1 4
5 5 -2
6 4 -4
2 5 8 9 15
*/

WA 10 pts。求助，嘎嘎挂分。

我的思路是对于每辆车处理出超速区间，然后区间按左端点排序，处理区间交。