[CSP-S 2024] 超速检测

这题写了一份WA40的代码，然后崩溃了/ll

一下是40分代码。

#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
struct node{int d, v, a;
}inp[N];
int T, n, m, L, V, p[N];
int main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);cin >> T;while (T--){cin >> n >> m >> L >> V;for (int i = 1; i <= n; i++){cin >> inp[i].d >> inp[i].v >> inp[i].a;}for (int i = 1; i <= m; i++){cin >> p[i];}set<int> st;int ans = 0;for (int i = 1; i <= n; i++){if (inp[i].a > 0 || inp[i].a == 0){if (inp[i].d > p[m]){continue;}if (inp[i].a == 0){if (inp[i].v > V){ans++;st.insert(p[m]);}continue;}int x = p[m] - inp[i].d;double tmp = sqrt(2.0 * inp[i].a * x + 1.0 * inp[i].v * inp[i].v);if (tmp > V * 1.0){ans++;st.insert(p[m]);}continue;}auto it = lower_bound(p + 1, p + m + 1, inp[i].d);int pos = *it;int x = pos - inp[i].d;double tmp2 = 2.0 * inp[i].a * x + 1.0 * inp[i].v * inp[i].v;if (tmp2 <= 0){continue;}else{if (sqrt(tmp2) > V){ans++;st.insert(pos);}}}cout << ans << " " << m - st.size() << '\n';}
}

发现错误的朋友可以在评论区里指出来。

问题就在于，如果我有一辆车减速，而在经过多个测速点时都超速，从极限的思想来说，我们只保留第一个即可。貌似是对的，其实是错的。

如果是这样，那么如果存在另一辆车，也经过了这些测速点，但他是一辆加速的车，那么我们会保留最后一个。

实际上，我们只需要保留一个，但保留了两个，故这样是错的。

所以我们就要对于每一辆车有一个检测区间，问题转化为区间覆盖问题。

OK，上正解！

#include<bits/stdc++.h>
using namespace std;
const int N = 100005;
int n,m,L,V;
struct node{int x, y;bool operator <(const node &q) const{return (y == q.y ? x < q.x : y < q.y);}
};
vector<node> ve;
struct cs{int d, v, a;
}cr[N];
int dct[N];
int main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int t;cin >> t;while (t--){cin >> n >> m >> L >> V;for (int i = 1; i <= n; i++) cin >> cr[i].d >> cr[i].v >> cr[i].a;for (int i = 1; i <= m; i++) cin >> dct[i];for (int i = 1; i <= n; i++){int st = -1, ed = -1;if (cr[i].a == 0){if (cr[i].v > V){st = lower_bound(dct + 1, dct + 1 + m, cr[i].d) - dct;ed = m;}}else if (cr[i].a > 0){if (cr[i].v > V){st = lower_bound(dct + 1, dct + 1 + m, cr[i].d) - dct;ed = m;}else if (cr[i].v == V){st = upper_bound(dct + 1, dct + 1 + m, cr[i].d) - dct;ed = m;}else{int fz= V * V - cr[i].v * cr[i].v;int fm = 2 * cr[i].a;st = upper_bound(dct + 1, dct + 1 + m, cr[i].d + fz / fm) - dct;ed = m;}}else{if (cr[i].v > V){int fz = cr[i].v * cr[i].v - V * V;int fm = -2 * cr[i].a;st = lower_bound(dct + 1, dct + 1 + m, cr[i].d) - dct;if (fz % fm == 0){ed = lower_bound(dct + 1, dct + 1 + m, cr[i].d + fz / fm) - dct - 1;}else{ed = lower_bound(dct + 1, dct + 1 + m, cr[i].d + fz / fm + 1) - dct - 1;}}}if (st != -1 && st <= ed) ve.push_back((node){st,ed});}sort(ve.begin(), ve.end());int nr = -0x3f3f3f3f, nl = 0, ans=0;for (int i = 0; i < ve.size(); i++){nl = max(nl, ve[i].x);if (nl > nr) ans++, nr = ve[i].y, nl = ve[i].x;}cout << ve.size() << " " << m - ans << "\n";ve.clear();}return 0;
}

OK，考场上也要多思考！！！