A little innovative.
#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<vector>#include<iostream>#define eps 0.000001#define maxn 5005using namespace std;int n;double w;struct node{ double x,y,d; bool operator<(const node &t)const { return d<t.d; }} no[maxn];int get(double x){ double ll=atan2(no[0].d,no[0].y-x),rr=atan2(no[0].d,no[0].x-x); for(int i=1;i<n;i++) { double l=atan2(no[i].d,no[i].y-x); double r=atan2(no[i].d,no[i].x-x);// printf("%lf %lf++\n",l,r); if(l-rr>eps)return 1; if(r-ll<-eps)return -1; ll=max(ll,l); rr=min(rr,r); } return 0;}bool solve(){ double l=0.0,r=w; while(r-l>eps) { double mid=(r+l)/2.0; int k=get(mid); if(k==0)return 1; else if(k==1)l=mid; else r=mid; } return 0;}int main(){ int t; scanf("%d",&t); while(t--) { scanf("%lf",&w); scanf("%d",&n); for(int i=0; i<n; i++) { scanf("%lf%lf%lf",&no[i].d,&no[i].x,&no[i].y); } sort(no,no+n); if(solve())puts("YES"); else puts("NO"); } return 0;}
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