1102 - 萌萌哒的第七题
Time Limit：
2s Memory Limit：
128MByte
Submissions：
371Solved：
38
DESCRIPTION
Feigay is doing some research about tornado. he drives
a car
and want
to reach
the center
of the tornado. The center
of tornado is (x0, y0)
and it is moving
at the speed is (vx,vy), Feigay is
at location (x1, y1) now, his velocity is v. He wants
to know whether he is possible
to reach that point.
INPUT
The
first line contains an integer T (<=
10000), indicates
the number of testcase.
Each test
case consists
of four lines,
the first line contains two integers x0,y0 ，
the second line contains two integers x1,y1,
the third line contains two integers vx,vy,
the fourth line contains one integer v.
The range
of every
number in the input
file is
0 ~
10000.
OUTPUT
For
each test
case, print
a line contains "YES" if it is possible
to reach
the center
or "NO" if impossible.
SAMPLE INPUT
3
0 0
1 1
1 0
1
0 0
1 1
2 0
1
0 0
2 1
2 0
1
SAMPLE OUTPUT
YES
NO
YES
SOLUTION
“玲珑杯”ACM比赛 Round
WA到爆 好好复习了一遍余弦定理：
cosα=b2+c2−a22bc
设
v0=|(vx,vy)|
为
(x0,y0)
的速度
(x0,y0)
到
(x1,y1)
的向量为
(x2,y2)
且
|(x2,y2)|=dis
α
为
(vx,vy)
与
(x2,y2)
的夹角 设
B⃗ =(vx,vy)
,
C⃗ =(x2,y2)
cosα=B⃗ ∗C⃗ |B⃗ ||C⃗ |
假设在t时间相遇 根据余弦定理: 设
|B⃗ |=v0t,|C⃗ |=dis,|A⃗ |=vt
cosα=|B⃗ |2|C⃗ |2−|A⃗ |22|B⃗ ||C⃗ |
得
(v20−v2)t2−2v0∗dis∗cosα∗t+dis2=0
判断t是否有>0的解即可
#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
#include<time.h>
#include<math.h>
#include<list>
#include<cstring>
#include<fstream>
#include<queue>
#include<sstream>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define INF 1000000007
#define pll pair<ll,ll>
#define pid pair<int,double>
const int inf=
1e9+
7;
const double EPS=
1e-9;
bool slove(
int x0,
int y0,
int x1,
int y1,
int vx,
int vy,
int v){
int v0_2=vx*vx+vy*vy;
double v0=
sqrt(v0_2);
int x2=x1-x0,y2=y1-y0;
int dis_2=x2*x2+y2*y2;
double dis=
sqrt(dis_2);
double cosT=(x2*vx+y2*vy)/(dis*v0);
double a=v0_2-v*v;
double b=-
2.0*v0*dis*cosT;
double c=dis_2;
double tmp=b*b-a*c*
4.0;
if(tmp<
0){
return false;
}
double ans1=(-b+
sqrt(tmp))/(
2*a);
double ans2=(-b-
sqrt(tmp))/(
2*a);
return ans1>
0||ans2>
0;
}
int main()
{
int T;
int x0,y0,x1,y1,vx,vy,v;
scanf(
"%d",&T);
while(T--){
scanf(
"%d%d%d%d%d%d%d",&x0,&y0,&x1,&y1,&vx,&vy,&v);
puts(slove(x0,y0,x1,y1,vx,vy,v)?
"YES":
"NO");
}
return 0;
}
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