ACM-Alice and Bob

题目描述：Alice and Bob

One day, Alice asks Bob to play a game called “K-in-a-row”. There is a game board whose size is N*M. Alice plays first, and they alternate in placing a piece of their color on an empty intersection. The winner is the first player to get an unbroken row of K stones horizontally, vertically, or diagonally. Now given the last situation of the game board, I would like to know who win or just a draw?

输入

The first line of input is the number of test cases T.

For each test case. The first line contains three integers N(3<= N <=15), M(3<=M<=15) and K(3<=K<=6).The next N line, each line contains M pieces, ‘A’ means Alice’s place and ‘B’ means Bob’s place while ‘O’ means empty. It is promised that at most one player wins.

输出

For each test case output the answer on a single line, if Alice wins then print “Alice Win!”, if Bob wins then print ”Bob Win!”, if no one wins, then print ”No Win!”.

样例输入

2
6 6 6
AOOOOO
BABBOO
OOAOBO
OOOAOO
OOBOAO
OOOOOA
5 5 3
AOBOA
BABAO
OOBOO
OOOOO
OOOOO

样例输出

Alice Win!
Bob Win!

思路：就是个五子棋，DFS即可。但是我的代码没有A，找不到问题。所以附上我的代码和正确代码。

我的代码：

// Alice and Bob_K_win.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"

//备注：没有AC！

#include <iostream>
#include <cstring>
using namespace std;

const int MAX = ;
int t, n, m, k,ans, vis[MAX][MAX],dir[][] = { , , -, , , , , -, , , -, -, , -, -,  };
char map[MAX][MAX];

void DFS(int x, int y,int a,int b)
{
    //cout << "x:" << x << "\ty:" << y << "\ta:" << a << "\tb:" << b << endl;

    if (ans != ) return;

    if (a == k || b == k)
    {
        if (a == k) ans = ;
        else ans = ;
        return;
    }

    for (int i = ; i < ; i++)
    {
        int nx = x + dir[i][];
        int ny = y + dir[i][];
        if (nx >=  && nx < n && ny >=  && ny < m && !vis[nx][ny] && map[nx][ny] != 'O')
        {
            //cout << "nx:" << nx << "\tny:" << ny << "\tmap[nx][ny]:" << map[nx][ny] << endl;
            vis[nx][ny] = ;
            if (map[nx][ny] == 'A') DFS(nx, ny, a + , b);
            else DFS(nx, ny, a, b + );
        }
    }

}

int main()
{
    cin >> t;
    while (t--)
    {
        memset(vis, , sizeof(vis));
        memset(map, '\0', sizeof(map));
        ans = ;

        cin >> n >> m >> k;
        for (int i = ; i < n; i++)
            cin >> map[i];

        for (int i = ; i < n; i++)
        {
            for (int j = ; j < m; j++)
            {
                if (map[i][j] != 'O' && !vis[i][j])
                {
                    vis[i][j] = ;
                    if (map[i][j] == 'A') DFS(i, j, , );
                    else if (map[i][j] == 'B') DFS(i, j, , );
                }

            }
        }

        if (ans == )
            cout << "Alice Win!" << endl;
        else if (ans == )
            cout << "Bob Win!" << endl;
        else
            cout << "No Win!" << endl;

    }

}

正确的代码：

#include<cstdio>
#include<iostream>
using namespace std;
int n,m,k,flag;
char map[][];
void dfs(int x,int y,int num,int dis,char e)
{
    if(num>k)
    {
        flag=;
        if(e=='A')
            cout<<"Alice Win!"<<endl;
        else
            cout<<"Bob Win!"<<endl;
        return ;
    }
    if(map[x][y]==e && x<n && x>= && y<m && y>=)
    {
        switch(dis)
        {
        case : dfs(x,y+,num+,dis,e);break;
        case : dfs(x,y-,num+,dis,e);break;
        case : dfs(x+,y,num+,dis,e);break;
        case : dfs(x-,y,num+,dis,e);break;
        case : dfs(x+,y+,num+,dis,e);break;
        case : dfs(x+,y-,num+,dis,e);break;
        case : dfs(x-,y-,num+,dis,e);break;
        case : dfs(x-,y+,num+,dis,e);break;
        }
    }
}
void solve()
{
    flag=;
    for(int i=;i<n;i++)//A
    {
        for(int j=;j<m;j++)
        {
            if(map[i][j]=='A')
            {
                for(int k=;k<=;k++)
                    dfs(i,j,,k,'A');
            }
            if(flag)
                break;
        }
        if(flag)
            break;
    }
    if(!flag)
    {
        for(int i=;i<n;i++)//B
        {
            for(int j=;j<m;j++)
            {
                if(map[i][j]=='B')
                {
                    for(int k=;k<=;k++)
                        dfs(i,j,,k,'B');
                }
                if(flag)
                    break;
            }
            if(flag)
                break;
        }
    }
    if(!flag)
        cout<<"No Win!"<<endl;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        cin>>n>>m>>k;
        for(int i=;i<n;i++)
        {
            for(int j=;j<m;j++)
                cin>>map[i][j];
        }
        solve();
    }
    return ;
}