UVA11796- Dog Distance

標籤：style   blog   http   color   io   os   ar   strong   for   

題意是給出兩條軌跡，分別給出起點和終點，要求兩條軌跡同時開始跑，同時到達重點

問，中途兩點間最大距離和最小距離的差值

我的做法：

設一個速度，用向量法類比過程

My Code：

#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\n",TT,mx-mn);}}

原題：

 

C	Dog Distance
	Input	Standard Input
	Output	Standard Output

Two dogs, Ranga and Banga, are running randomly following two different paths. They both run forT seconds with different speeds. Ranga runs with a constant speed ofR 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} – {min (D(b)) 0 <= b <= T}

Given the paths of the two dogs, your job is to find the dog distance.

Each path will be represented usingN points, (P1 P2 P3 ... PN). The dog following this path will start from P1 and follow the line joining withP2, and then it will follow the line joiningP2-P3, thenP3-P4 and so on until it reachesPn.

Input

Input starts with an integer I(I≤1000), the number of test cases.

Each test case starts with 2 positive integersA(2≤A≤50),B(2≤B≤50). The next line contains the coordinates ofA points with the format X1Y1 X2 Y2 ...XA YA,(0≤Xi,Yi≤1000). These points indicate the path taken by Ranga. The next line containsB points in the same format. These points indicate the path taken by Banga. All distance units are given in meters and consecutive points are distinct. All the given coordinates are integers.

Note that the values ofT, R and S are unknown to us.

Output

For each case, output the case number first. Then output the dog distance rounded to the nearest integer. Look at the samples for exact format.

 

 

Sample Input	Sample Output
2 2 2 0 0 10 0 0 1 10 1 3 2 635 187 241 269 308 254 117 663 760 413



UVA11796- Dog Distance