Codeforces Round #327 (Div. 1) B题： Chip 'n Dale Rescue Rangers [数学题]

题意：一艘飞船最大驾驶速度为Vmax，现在点X1，Y1，想要飞到点X2，Y2，前T秒的有风速为VX,VY，T秒后的风速为WX,WY，问最短达到时间。（保证风速小于Vmax）

解法：一开始三分套二分，神奇的过了初测，但最后果然还是WA了（应该是我选取的角度有问题吧……），后来推了推公式就A了。只要将三段位移分开来考虑，最后行驶速度必然是Vmax才能最短时间到达，列个勾股定理的公式推导一下即可。

代码：

#include<stdio.h>#include<string.h>#include<algorithm>#include<math.h>#include<iostream>#include<stdlib.h>#include<set>#include<map>#include<queue>#include<vector>#include<bitset>#pragma comment(linker, "/STACK:1024000000,1024000000")template <class T>bool scanff(T &ret){ //Faster Input    char c; int sgn; T bit=0.1;    if(c=getchar(),c==EOF) return 0;    while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();    sgn=(c=='-')?-1:1;    ret=(c=='-')?0:(c-'0');    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');    if(c==' '||c=='\n'){ ret*=sgn; return 1; }    while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;    ret*=sgn;    return 1;}#define inf 1073741823#define llinf 4611686018427387903LL#define PI acos(-1.0)#define lth (th<<1)#define rth (th<<1|1)#define rep(i,a,b) for(int i=int(a);i<=int(b);i++)#define drep(i,a,b) for(int i=int(a);i>=int(b);i--)#define gson(i,root) for(int i=ptx[root];~i;i=ed[i].next)#define tdata int testnum;scanff(testnum);for(int cas=1;cas<=testnum;cas++)#define mem(x,val) memset(x,val,sizeof(x))#define mkp(a,b) make_pair(a,b)#define findx(x) lower_bound(b+1,b+1+bn,x)-b#define pb(x) push_back(x)using namespace std;typedef long long ll;typedef pair<int,int> pii;double ox1,oy1,ox2,oy2;double vx1,vy1,vx2,vy2;double vmax,t;double dx,dy;double eps=1e-9;double gettime1(double vx,double vy,double ox,double oy){    double lenx=ox2-ox;    double leny=oy2-oy;    double a=vx*vx+vy*vy-vmax*vmax;    double b=-2.0*(vx*lenx+vy*leny);    double c=(lenx*lenx+leny*leny);    double x1= (-b-sqrt(b*b-4.0*a*c))/a/2.0;    double x2= (-b+sqrt(b*b-4.0*a*c))/a/2.0;    if(x1>-eps)return x1;    return x2;}double gettime2(double vx,double vy,double ox,double oy){    double lenx=ox2-ox;    double leny=oy2-oy;    double a=vx*vx+vy*vy-vmax*vmax;    double b=-2.0*(vx*(lenx+t*vx)+vy*(leny+t*vy));    double c=(lenx+t*vx)*(lenx+t*vx)+(leny+t*vy)*(leny+t*vy);    double x1= (-b-sqrt(b*b-4.0*a*c))/a/2.0;    double x2= (-b+sqrt(b*b-4.0*a*c))/a/2.0;    if(x1>-eps)return x1;    return x2;}int main(){    scanff(ox1);scanff(oy1);    scanff(ox2);scanff(oy2);    scanff(vmax);scanff(t);    scanff(vx1);scanff(vy1);    scanff(vx2);scanff(vy2);    double t1=gettime1(vx1,vy1,ox1,oy1);    double t2=gettime2(vx2,vy2,ox1+vx1*t,oy1+vy1*t);    if(t1<t)printf("%.15f\n",t1);    else printf("%.15f\n",t2);    return 0;}