题意:y=ax+b的直线上有n个点,给出这n个点的(x[i],vx[i],vy[i]),(vx[i],vy[i]为第i个点x,y的速度)
当两个点碰撞时,总价值+=2.n<=2e5,-1e9<=x[i]<=1e9.问总价值为多少?
若(x1,vx1,vy1),(x2,vx2,vy2)能在t秒相遇
x1+t*vx[1]= x2+t*vx[2]
y1+t*vy[1]= y2+t*vy[2]
则 (x2-x1)/(vx[1]-vx[2]) = (y2-y1)/(vy[1]-vy[2])
y1,y2用ax+b代入得到. a*(v[x1]-v[x2]) = vy[1]-vy[2]
最后两点能相遇 只有a*vx[i]-vy[i] = a*vx[j]-v[j].
用map记录,k=mp[i] 表示这有k个点能在同一时刻相遇:C(k,2)*2即为其贡献.
当两个点碰撞时,总价值+=2.n<=2e5,-1e9<=x[i]<=1e9.问总价值为多少?
若(x1,vx1,vy1),(x2,vx2,vy2)能在t秒相遇
x1+t*vx[1]= x2+t*vx[2]
y1+t*vy[1]= y2+t*vy[2]
则 (x2-x1)/(vx[1]-vx[2]) = (y2-y1)/(vy[1]-vy[2])
y1,y2用ax+b代入得到. a*(v[x1]-v[x2]) = vy[1]-vy[2]
最后两点能相遇 只有a*vx[i]-vy[i] = a*vx[j]-v[j].
用map记录,k=mp[i] 表示这有k个点能在同一时刻相遇:C(k,2)*2即为其贡献.
同样去除掉(vx,vy)相同方案.
#include <bits/stdc++.h> using namespace std; typedef long long ll; typedef pair<ll,ll> ii; const int N=3e5+5; ll n,a,b,x[N],vx[N],vy[N]; map<ll,ll> mp; map<ii,ll> mk; int main() { ios::sync_with_stdio(false); cin.tie(0); cin>>n>>a>>b; for(int i=1;i<=n;i++) { cin>>x[i]>>vx[i]>>vy[i]; mp[a*vx[i]-vy[i]]++; mk[ii(vx[i],vy[i])]++; } ll res=0; for(auto it=mp.begin();it!=mp.end();it++) { ll val=it->second; res+=val*(val-1); } for(auto it=mk.begin();it!=mk.end();it++) { ll val=it->second; res-=val*(val-1); } cout<<res<<'\n'; return 0; }