hdu5299 Circles Game

Circles Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1427    Accepted Submission(s): 451

Problem Description

There are n circles on a infinitely large table.With every two circle, either one contains another or isolates from the other.They are never crossed nor tangent.

Alice and Bob are playing a game concerning these circles.They take turn to play,Alice goes first:

1、Pick out a certain circle A,then delete A and every circle that is inside of A.

2、Failling to find a deletable circle within one round will lost the game.

Now,Alice and Bob are both smart guys,who will win the game,output the winner's name.

 

Input

The first line include a positive integer T<=20,indicating the total group number of the statistic.

As for the following T groups of statistic,the first line of every group must include a positive integer n to define the number of the circles.

And the following lines,each line consists of 3 integers x,y and r,stating the coordinate of the circle center and radius of the circle respectively.

n≤20000。|x|≤20000。|y|≤20000，r≤20000。

 

Output

If Alice won,output “Alice”,else output “Bob”

 

Sample Input

2
1
0 0 1
6
-100 0 90
-50 0 1
-20 0 1
100 0 90
47 0 1
23 0 1

 

Sample Output

Alice
Bob

 

Author

FZUACM

 

Source

2015 Multi-University Training Contest 1

显然圆的包括关系能够构成一片森林，然后问题就能够转化为：每一步能够删除森林的一棵树或者某树的一棵子树。不能删者输。

这样。问题就变成经典的树上删边游戏了。

树上删边游戏有一个非常重要的结论：叶子节点的SG值为0，中间节点的SG值为它的全部子节点的SG值加1后的异或和。(证明详见贾志豪论文《组合游戏略述——浅谈SG游戏的若干拓展及变形》)

如今。我们的主要问题就是怎样把圆的包括关系转化为森林。这里要用到圆的扫描线算法。首先对于每一个圆。创建两个时间点，一个进入一个离开。再对全部时间点从小到大排序。

然后逐个处理时间点。用set维护全部圆。每遇到一个进入时间。分情况讨论圆的位置关系。然后把这个圆插入set中。每遇到一个离开时间，从set中删除这个圆。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<set>
#define F(i,j,n) for(int i=j;i<=n;i++)
#define D(i,j,n) for(int i=j;i>=n;i--)
#define ll long long
#define maxn 20010
using namespace std;
int t,n,cnt,tt,tmp;
int head[maxn],fa[maxn];
struct cir{int x,y,r;}a[maxn];
struct edge_type{int next,to;}e[maxn*2];
struct bor
{
	int x,f,id;
	friend bool operator < (bor a,bor b)
	{
		if (a.x==b.x) return a.f<b.f;
		return a.x<b.x;
	}
}q[maxn*2];
double get_h(int id,int x,int opt)
{
	return a[id].y+opt*sqrt(a[id].r*a[id].r-(a[id].x-x)*(a[id].x-x));
}
struct pos
{
	int id,opt;
	pos(int a=0,int b=0){id=a;opt=b;}
	friend bool operator < (pos a,pos b)
	{
		if (a.id==b.id) return a.opt<b.opt;
		return get_h(a.id,tmp,a.opt)<get_h(b.id,tmp,b.opt);
	}
};
set<pos> s;
set<pos>::iterator it;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
	while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
inline void add_edge(int x,int y)
{
	e[++cnt]=(edge_type){head[x],y};
	head[x]=cnt;
	fa[y]=x;
}
inline ll get(int x)
{
	ll ret=0;
	for(int i=head[x];i;i=e[i].next) ret^=get(e[i].to);
	return ret+1;
}
int main()
{
	t=read();
	while (t--)
	{
		cnt=tt=0;
		memset(fa,0,sizeof(fa));
		memset(head,0,sizeof(head));
		n=read();
		F(i,1,n)
		{
			a[i].x=read();a[i].y=read();a[i].r=read();
			q[++tt]=(bor){a[i].x-a[i].r,1,i};
			q[++tt]=(bor){a[i].x+a[i].r,-1,i};
		}
		sort(q+1,q+tt+1);
		F(i,1,tt)
		{
			if (q[i].f==1)
			{
				tmp=q[i].x;
				it=s.lower_bound(pos(q[i].id,1));
				if (it!=s.end())
				{
					if (it->opt==1) add_edge(it->id,q[i].id);
					else if (fa[it->id]) add_edge(fa[it->id],q[i].id);
				}
				s.insert(pos(q[i].id,1));
				s.insert(pos(q[i].id,-1));
			}
			else
			{
				s.erase(pos(q[i].id,1));
				s.erase(pos(q[i].id,-1));
			}
		}
		ll sum=0;
		F(i,1,n) if (!fa[i]) sum=sum^get(i);
		puts(sum?"Alice":"Bob");
	}
}

阅读全文

顶

1

踩

0

• 上一篇bzoj2049【SDOI2008】Cave 洞穴勘測
• 下一篇bzoj3144【HNOI2013】切糕