Codeforces 793C Mice Problem 几何(模拟)

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题意:n个点(x,y),速度(vx,vy),n<=1e5,问n个点第一次同时落在矩形内时间?

某个点落在矩形内的时间t为:x1<=x+t1*vx<=x2,(x1-x)/vx<=t1<=(x2-x)/vx,y1<=y+t2*vy<=y2 (y1-y)/vy<=t2<=(y2-y)/vy

max(t1,t2)<=t<=min(t1,t2) 每个点落在矩形内的时间看成一条线段,则求n条线段公共部分的左端点即可

#include <bits/stdc++.h>using namespace std;typedef long long ll;typedef pair<ll,ll> ii;const int N=1e3+20;const int inf=2e8;const double eps=1e-11;int n;int main(){double x1,y1,x2,y2;double x,y,vx,vy;double L=0,R=1e9;while(cin>>n){bool flag=true;scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);for(int i=0;i<n;i++){double lx,rx,ly,ry,l,r;scanf("%lf%lf%lf%lf",&x,&y,&vx,&vy);if(vx==0)lx=0,rx=(x>x1&&x<x2)?inf:-1;else{lx=(x1-x)/vx,rx=(x2-x)/vx;if(vx<0)swap(lx,rx);}if(vy==0)ly=0,ry=(y>y1&&y<y2)?inf:-1;else{ly=(y1-y)/vy,ry=(y2-y)/vy;if(vy<0)swap(ly,ry);}if(ly>ry||lx>rx)flag=false;//else{l=max(lx,ly);r=min(rx,ry);//l<=t<=rL=max(l,L);R=min(r,R);}if(L>R)flag=false;}if(R-L>=eps&&flag)printf("%.10lf\n",L);elseputs("-1");}return 0;}

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