Circles Game
题目连接：
http://acm.hdu.edu.cn/showproblem.php?pid=5299
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
Hint
题意
有一个平面，平面上有n个圆，圆与圆之间只有相离和包含的关系。
然后他们开始跑博弈论。
每个人可以拿圆和他所包含的圆。
如果谁不能拿了的话，那就输了。
问你谁能赢。
题解：
很显然，包含和相离关系，那么就是一个森林之间的关系。
然后把这个森林需要建立出来，这个就直接暴力就好了。
然后后面的博弈是一个很经典的东西。
叶子节点的SG值为0；中间节点的SG值为它的所有子节点的SG值加1 后的异或和。
然后直接跑就好了。
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+6;
vector<int> E[maxn];
struct node
{
int x,y,r;
};
node a[maxn];
int dp[maxn];
bool cmp(node A,node B)
{
return A.r>B.r;
}
int fa[maxn];
bool check(node A,node B)
{
long long k = 1ll*(A.x-B.x)*(A.x-B.x)+1ll*(A.y-B.y)*(A.y-B.y);
return k<=1ll*A.r*A.r;
}
void init()
{
memset(a,0,sizeof(a));
memset(fa,0,sizeof(fa));
memset(dp,0,sizeof(dp));
for(int i=0;i<maxn;i++)E[i].clear();
}
void dfs(int x,int fa)
{
dp[x]=0;
for(int j=0;j<E[x].size();j++)
{
int v = E[x][j];
if(v==fa)continue;
dfs(v,x);
dp[x]^=(dp[v]+1);
}
}
void solve()
{
init();
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].r);
node k;
a[0].x=0,a[0].y=0,a[0].r=4e4+6;
sort(a+1,a+1+n,cmp);
for(int i=1;i<=n;i++)
{
for(int j=i-1;j>=0;j--)
if(check(a[j],a[i]))
{
fa[i]=j;
E[j].push_back(i);
break;
}
}
dfs(0,-1);
if(dp[0])cout<<"Alice"<<endl;
else cout<<"Bob"<<endl;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)solve();
}