UVA11796- Dog Distance
栏目：互联网时间：2014-09-30 07:55:45
题意是给出两条轨迹，分别给出起点和终点，要求两条轨迹同时开始跑，同时到达重点
问，中途两点间最大距离和最小距离的差值
我的做法：
设一个速度，用向量法模拟过程
我的代码：
#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
const double inf=1e9;
struct dot
{
double x,y;
dot(){}
dot(double a,double b){x=a;y=b;}
dot operator +(dot a){return dot(x+a.x,y+a.y);}
dot operator -(dot a){return dot(x-a.x,y-a.y);}
dot operator *(double a){return dot(x*a,y*a);}
double operator *(dot a){return x*a.y-y*a.x;}
dot operator /(double a){return dot(x/a,y/a);}
double operator /(dot a){return x*a.x+y*a.y;}
bool operator ==(dot a){return x==a.x&&y==a.y;}
void in(){scanf("%lf%lf",&x,&y);}
void out(){printf("%lf %lf",x,y);}
double mod(){return sqrt(x*x+y*y);}
double dis(dot a){return sqrt(pow(x-a.x,2)+pow(y-a.y,2));}
};
double mxd(dot a,dot b,dot c)
{
return max(a.dis(b),a.dis(c));
}
double mnd(dot a,dot b,dot c)
{
if((a-b)/(c-b)<=0)
return a.dis(b);
if((a-c)/(b-c)<=0)
return a.dis(c);
return fabs((a-b)*(c-b)/b.dis(c));
}
int main()
{
double va,vb,mx,mn;
dot a[110],b[110],sa,sb,s,e,ta,tb,v,t;
int i,j,T,TT,n,m;
cin>>T;
for(TT=1;TT<=T;TT++)
{
mx=0;
mn=inf;
cin>>n>>m;
for(i=0;i<n;i++)
a[i].in();
va=0;
for(i=1;i<n;i++)
va+=a[i-1].dis(a[i]);
for(i=0;i<m;i++)
b[i].in();
vb=0;
for(i=1;i<m;i++)
vb+=b[i-1].dis(b[i]);
sa=a[0];sb=b[0];
i=j=1;
while(i<n&&j<m)
{
ta=a[i]-sa;
ta=ta/ta.mod();
ta=ta*va;
tb=b[j]-sb;
tb=tb/tb.mod();
tb=tb*vb;
if(sa.dis(a[i])/va>sb.dis(b[j])/vb)
{
v=tb-ta;
t=sa;
s=sb;
e=s+v*(sb.dis(b[j])/vb);
sa=sa+ta*(sb.dis(b[j])/vb);
sb=b[j++];
}
else
{
v=ta-tb;
t=sb;
s=sa;
e=s+v*(sa.dis(a[i])/va);
sb=sb+tb*(sa.dis(a[i])/va);
sa=a[i++];
}
mx=max(mx,mxd(t,s,e));
mn=min(mn,mnd(t,s,e));
}
printf("Case %d: %.0lf
",TT,mx-mn);
}
}
原题：
|
C |
Dog Distance |
|
|
Input |
Standard Input |
|
|
Output |
Standard Output |
|
Two dogs, Ranga and Banga, are running randomly following two different paths. They both run for T seconds with different speeds. Ranga runs with a constant speed of R m/s, whereas Banga runs with a constant speed of S m/s. Both the dogs start and stop at the same time. Let D(t) be the distance between the two dogs at time t.
The dog distance is equal to the difference between the maximum and the minimum distance between the two dogs in their whole journey.
Mathematically,
Dog Distance = {max (D(a)) 0 <= a <= T}
------分隔线----------------------------
下一篇 网站不添加内容也有好的排名
------分隔线----------------------------