2087.txt

北大ACM题目例程 详细的解答过程 程序实现 算法分析

#include"iostream.h"
#include"stdio.h"
#include"math.h"
#include<algorithm>
using namespace std;
/////////////////////////
#define Type double /*坐标类型*/
/////////////////////////
struct point
{
	Type x,y;
	int host;
};
const double pi=3.1415926535898;
inline Type cheng(point a,point b,point c)
{return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);}
inline Type dcheng(point a,point b,point c)
{return (b.x-a.x)*(c.x-a.x)+(c.y-a.y)*(b.y-a.y);}
double jl(point a,point b)
{
	return ((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
const double eps=0.0000001;
point object(point a,double degree,double len)
{
	point c;
	c.x=a.x+len*cos(degree/180*pi);
	c.y=a.y+len*sin(degree/180*pi);
	c.host=a.host;

	return c;
}
point p[10];
int n;
void throwball(point a,point b,int prev)
{
	point pt;
	int k=-1,i;double l=9999999,t;
	for(i=0;i<n;i++)
	if(i!=prev)
	{
		if(fabs(cheng(p[i],a,b))<eps&&dcheng(p[i],a,b)<-eps)
		{
			t=jl(p[i],a);
			
			if(t<l)
			{
				k=i;
				l=t;
			}
		}
	}
	if(k<0)
	{
		p[n++]=b;
		for(i=0;i<n;i++)
		if(p[i].host==-1)break;
		
		if(i!=0)swap(p[i],p[0]);
	}
	else 
	{
		pt=p[k];
		p[k].host=a.host;
		b.host=pt.host;
		throwball(pt,b,k);
	}
}
int cmp(point a,point b)
{
	return jl(a,p[0])<jl(b,p[0]);
}
void doit()
{
	int key;int use[2]={0,0};
	char name1[100],name2[100];
	cin>>name1>>name2;
	point start;
	double degree,len,t;
	cin>>start.x>>start.y>>degree>>len;
	p[0]=object(start,degree,len);
	p[0].host=-1;
	n=1;
	cin>>start.x>>start.y>>degree>>len;
	start.host=0;
	throwball(start,object(start,degree,len),-1);
	cin>>start.x>>start.y>>degree>>len;
	start.host=1;
	throwball(start,object(start,degree,len),-1);
	int i,j,k;
	for(i=3;i<7;i++)
	{
		k=-1;len=9999999;
		if(use[0]==2)key=0;
		else if(use[1]==2)key=1;
		else for(j=1;j<n;j++)
		{
			t=jl(p[0],p[j]);
			if(k==-1||t<len)
			{
				k=j;
				len=t;
				key=p[j].host;
			}
		}		
		cin>>start.x>>start.y>>degree>>len;
		start.host=1-key;
		if(use[start.host]>=2)while(1);
		use[start.host]++;

		throwball(start,object(start,degree,len),-1);
	}
	i=i;
	sort(&p[1],&p[7],cmp);
	if(p[1].host==0)cout<<name1<<' ';
	else cout<<name2<<' ';
	for(i=1;p[i].host==p[1].host;i++)		;
	cout<<i-1<<endl;
	return;
}
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		doit();
	}
	return 0;
}