Joy of Flight 2014-2015 ACM-ICPC, NEERC, Northern Subregional Contest
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题意:
给出风速,最大速度,起点终点,询问是否能到达终点,以及没秒所在位置。
思路:
风速的矢量分离出来,求对于位移的影响,作用在dx dy上,求平均X的速度和Y的速度
#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<vector>#include <map>#include <queue>using namespace std;typedef long long ll;double t[100005];double vx[10005];double vy[10005];int main(){ freopen("joy.in","r",stdin); freopen("joy.out","w",stdout); double sx,sy,fx,fy; double n,k,v; cin>>sx>>sy>>fx>>fy>>n>>k>>v; for(int i=1;i<=n;i++) cin>>t[i]>>vx[i]>>vy[i]; t[(int)n+1]=k; vx[(int)n+1]=vy[(int)n+1]=0; double lx=0,ly=0; for(int i=1;i<=n;i++) { lx+=(vx[i] )*(t[i+1]-t[i]); ly+=(vy[i] )*(t[i+1]-t[i]); } double dx=-lx+fx-sx,dy=-ly+fy-sy; double px=(dx)/k,py=(dy)/k; if(dx*dx+dy*dy>v*v*k*k) { printf("No\n"); return 0; } printf("Yes\n"); int cur=0; double nowfx=0,nowfy=0; for(int i=0;i<k;i++) { if(i==t[cur+1]) nowfx=vx[++cur],nowfy=vy[cur]; sx+=px+nowfx; sy+=py+nowfy; printf("%.6lf %.6lf\n",sx,sy); } return 0;}
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