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



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
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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