hdu 5299 Circles Game
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Circles Game
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 547 Accepted Submission(s): 150
Total Submission(s): 547 Accepted Submission(s): 150
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
210 0 16-100 0 90-50 0 1-20 0 1100 0 9047 0 123 0 1
Sample Output
AliceBob
题意:在坐标系中给出一些圆,这些圆可能相互包含和相离,现在两个人玩一个游戏,每个人每次选择去掉一个圆,如果这个圆中有小圆,同时去掉,没有圆可取的人输。
分析:等级制的威佐夫博奕
因为圆与圆有相互包含的关系,那么我们可以建一棵树,虚设一个无穷大的圆作为根节点,那么剩下的圆与大圆建立联系。
比如例2的树:
<span style="font-size:18px;"> 0 / \ 1 4 / \ / \ 2 3 5 6</span>
那么每一个子树进行威佐夫博奕,将结果传给父节点继续博弈,最终大圆的根节点博弈结果就是答案
<span style="font-size:18px;">#include<cstring>#include<string>#include<iostream>#include<queue>#include<cstdio>#include<algorithm>#include<map>#include<cstdlib>#include<cmath>#include<vector>using namespace std;#define INF 0x3f3f3f3fstruct node{ int x,y; int r; int val; bool operator < (const node &A) const { return r>A.r; }} T[20004];vector<int>vec[20004];void init(int n){ for(int i=0; i<=n; i++) { vec[i].clear(); } T[0].x=0; T[0].y=0; T[0].r=100000; T[0].val=0;}bool include(int k,int num){ long long dis=(T[k].x-T[num].x)*(T[k].x-T[num].x)+(T[k].y-T[num].y)*(T[k].y-T[num].y); if(dis<=T[k].r*T[k].r) return true; return false;}void dfs_find(int k,int num){ for(int i=0; i<vec[k].size(); i++) { if(include(vec[k][i],num)) { dfs_find(vec[k][i],num); return ; } } vec[k].push_back(num); return ;}int dfs_solve(int k){ int ans=0; for(int i=0;i<vec[k].size();i++) { ans^=dfs_solve(vec[k][i]); } T[k].val+=ans; return T[k].val;}int main(){ int T_T; scanf("%d",&T_T); while(T_T--) { int n; scanf("%d",&n); init(n); for(int i=1; i<=n; i++) { scanf("%d%d%d",&T[i].x,&T[i].y,&T[i].r); T[i].val=1; } sort(T+1,T+n+1); for(int i=1; i<=n; i++) { dfs_find(0,i); } dfs_solve(0); if(T[0].val!=0) printf("Alice\n"); else printf("Bob\n"); } return 0;}</span>
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