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();
}