Codeforces 924D Contact ATC (看题解)

Contact ATC

我跑去列方程， 然后就gg了。。。

我们计每个飞机最早到达时间为L[ i ], 最晚到达时间为R[ i ]， 

对于面对面飞行的一对飞机， 只要他们的时间有交集则必定满足条件。

对于相同方向飞行的飞机， 只有其中一个的时间包含另一个的时间才满足条件。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long longusing namespace std;const int N = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);struct Bit {int a[N];void init() {memset(a, 0, sizeof(a));}void modify(int x, int v) {for(int i = x; i < N; i += i & -i)a[i] += v;}int sum(int x) {int ans = 0;for(int i = x; i; i -= i & -i)ans += a[i];return ans;}int query(int L, int R) {if(L > R) return 0;return sum(R) - sum(L - 1);}
};struct Node {Node(LL a, LL b) : a(a), b(b) {}bool operator < (const Node& rhs) const {return a * rhs.b < rhs.a * b;}bool operator == (const Node& rhs) const {return a * rhs.b == rhs.a * b;}void print() {printf("%.5f ", 1.0 * a / b);}LL a, b;
};int n, w, x[N], v[N];
LL ans = 0;
vector<PII> vc[2];
vector<Node> hs;
Bit bit;bool cmp(PII& a, PII& b) {if(a.fi == b.fi) return a.se < b.se;return a.fi > b.fi;
}LL solve(vector<PII>& vc) {bit.init();LL ans = 0;sort(vc.begin(), vc.end(), cmp);for(int i = 0; i < SZ(vc); i++) {ans += bit.sum(vc[i].se);bit.modify(vc[i].se, 1);}return ans;
}int main() {scanf("%d%d", &n, &w);for(int i = 1; i <= n; i++) {scanf("%d%d", &x[i], &v[i]);if(x[i] < 0) {hs.push_back(Node(-x[i], v[i] + w));hs.push_back(Node(-x[i], v[i] - w));} else {hs.push_back(Node(x[i], w - v[i]));hs.push_back(Node(x[i], -w - v[i]));}}sort(hs.begin(), hs.end());hs.erase(unique(hs.begin(), hs.end()), hs.end());for(int i = 1; i <= n; i++) {if(x[i] < 0) {int L = lower_bound(hs.begin(), hs.end(), Node(-x[i], v[i] + w)) - hs.begin() + 1;int R = lower_bound(hs.begin(), hs.end(), Node(-x[i], v[i] - w)) - hs.begin() + 1;vc[0].push_back(mk(L, R));} else {int L = lower_bound(hs.begin(), hs.end(), Node(x[i], w - v[i])) - hs.begin() + 1;int R = lower_bound(hs.begin(), hs.end(), Node(x[i], -w - v[i])) - hs.begin() + 1;vc[1].push_back(mk(L, R));}}ans += solve(vc[0]);ans += solve(vc[1]);ans += 1ll * SZ(vc[0]) * SZ(vc[1]);bit.init();for(auto& t : vc[1]) bit.modify(t.se, 1);for(auto& t : vc[0]) ans -= bit.sum(t.fi - 1);bit.init();for(auto& t : vc[1]) bit.modify(t.fi, 1);for(auto& t : vc[0]) ans -= bit.query(t.se + 1, N - 1);printf("%lld\n", ans);return 0;
}/*
*/