coderforce 69D.Dot

最新推荐文章于 2022-01-30 01:41:42 发布

Greaterlee

最新推荐文章于 2022-01-30 01:41:42 发布

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分类专栏：博弈搜索文章标签： coderforce 69D Dot 记忆化搜索博弈入门

本文链接：https://blog.youkuaiyun.com/Greaterlee/article/details/52172176

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本文介绍了一个基于坐标平移和反射的两人回合制游戏，通过深度优先搜索算法确定初始状态下哪位玩家能赢得游戏。游戏规则包括移动点的位置及对角线反射，并限制了超出特定距离的移动。

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Anton and Dasha like to play different games during breaks on checkered paper. By the 11th grade they managed to play all the games of this type and asked Vova the programmer to come up with a new game. Vova suggested to them to play a game under the code name "dot" with the following rules:

On the checkered paper a coordinate system is drawn. A dot is initially put in the position (x, y).
A move is shifting a dot to one of the pre-selected vectors. Also each player can once per game symmetrically reflect a dot relatively to the line y = x.
Anton and Dasha take turns. Anton goes first.
The player after whose move the distance from the dot to the coordinates' origin exceeds d, loses.

Help them to determine the winner.

Input

The first line of the input file contains 4 integers x, y, n, d ( - 200 ≤ x, y ≤ 200, 1 ≤ d ≤ 200, 1 ≤ n ≤ 20) — the initial coordinates of the dot, the distance d and the number of vectors. It is guaranteed that the initial dot is at the distance less than d from the origin of the coordinates. The following n lines each contain two non-negative numbers x_i and y_i (0 ≤ x_i, y_i ≤ 200) — the coordinates of the i-th vector. It is guaranteed that all the vectors are nonzero and different.

Output

You should print "Anton", if the winner is Anton in case of both players play the game optimally, and "Dasha" otherwise.

Sample Input

DFS记录路径的博弈，很好的一道题。

Input

0 0 2 3
1 1
1 2

Output

Anton

Input

0 0 2 4
1 1
1 2

Output

Dasha

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int mapp[500][500];
int a[300],b[300],bx,by,d,n;
int dfs(int x,int y)
{
    if((x-250)*(x-250)+(y-250)*(y-250)>=d*d)
        return 1;
    if(mapp[x][y]!=-1)
        return mapp[x][y];
    for(int i=0; i<n; i++)
    {
        if(dfs(x+a[i],y+b[i])==0)
            return mapp[x][y]=1;
    }
    return mapp[x][y]=0;
}
int main()
{
    cin>>bx>>by>>n>>d;
    bx+=250;
    by+=250;
    for(int i=0; i<n; i++)
        scanf("%d%d",&a[i],&b[i]);
    memset(mapp,-1,sizeof(mapp));
    if(dfs(bx,by))
        printf("Anton\n");
    else
        printf("Dasha\n");
    return 0;
}