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
     
      #define LL long long#define fi first#define se second#define mk make_pair#define PLL pair
      
       #define PLI pair
       
        #define PII pair
        
         #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
         
           vc[2];vector
          
            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
           
            & 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;}/**/