题意是这样。给出非常多圆，要么两两相离，要么包括，若删掉一个圆，那被他包括的都要删除，若某人没有圆给他删，那么他就赢了。
。。。知道树上博弈的话。就非常easy。
。。不知道的话。这确实是个神题……
按半径上升排序，从左往右扫。i扫到第一个j能够包括它的圆，建立j到i的连边，然后break
这样就建立好了一棵树，之后知道这个就非常easy了。。。
树的删边游戏
规则例如以下:
给出一个有 N 个点的树,有一个点作为树的根节点。
游戏者轮流从树中删去边,删去一条边后,不与根节点相连的
部分将被移走。
谁无路可走谁输。
我们有例如以下定理:
[定理]
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。
当然啦，像我这样暴力的写法。交c++会超时#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
struct Point
{
int x,y,r;
}point[20010];
bool cmp(Point a,Point b)
{
return a.r<b.r;
}
struct Edge
{
int to,next;
}edge[20010];
int head[20010],tail;
void add(int from,int to)
{
edge[tail].to=to;
edge[tail].next=head[from];
head[from]=tail++;
}
int dfs(int from)
{
int ans=0;
for(int i=head[from];i!=-1;i=edge[i].next)
ans^=dfs(edge[i].to)+1;
return ans;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d%d%d",&point[i].x,&point[i].y,&point[i].r);
sort(point,point+n,cmp);
tail=0;
memset(head,-1,sizeof(head));
for(int i=0;i<n;i++)
{
bool flag=0;
for(int j=i+1;j<n;j++)
if(ll(point[j].r*point[j].r)>ll(point[i].x-point[j].x)*(point[i].x-point[j].x)+(point[i].y-point[j].y)*(point[i].y-point[j].y))
{
flag=1;
add(j,i);
break;
}
if(!flag)
add(n,i);
}
if(dfs(n)!=0)
puts("Alice");
else
puts("Bob");
}
}Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 593 Accepted Submission(s): 164
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.
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。
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