Find the Winning Move

Description

4x4 tic-tac-toe is played on a board with four rows (numbered 0 to 3 from top to bottom) and four columns (numbered 0 to 3 from left to right). There are two players, x and o, who move alternately with x always going first. The game is won by the first player to get four of his or her pieces on the same row, column, or diagonal. If the board is full and neither player has won then the game is a draw.  Assuming that it is x's turn to move, x is said to have a forced win if x can make a move such that no matter what moves o makes for the rest of the game, x can win. This does not necessarily mean that x will win on the very next move, although that is a possibility. It means that x has a winning strategy that will guarantee an eventual victory regardless of what o does.  Your job is to write a program that, given a partially-completed game with x to move next, will determine whether x has a forced win. You can assume that each player has made at least two moves, that the game has not already been won by either player, and that the board is not full. 

Input

The input contains one or more test cases, followed by a line beginning with a dollar sign that signals the end of the file. Each test case begins with a line containing a question mark and is followed by four lines representing the board; formatting is exactly as shown in the example. The characters used in a board description are the period (representing an empty space), lowercase x, and lowercase o. For each test case, output a line containing the (row, column) position of the first forced win for x, or '#####' if there is no forced win. Format the output exactly as shown in the example.

Output

For this problem, the first forced win is determined by board position, not the number of moves required for victory. Search for a forced win by examining positions (0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), ..., (3, 2), (3, 3), in that order, and output the first forced win you find. In the second test case below, note that x could win immediately by playing at (0, 3) or (2, 0), but playing at (0, 1) will still ensure victory (although it unnecessarily delays it), and position (0, 1) comes first.

Sample Input

?

....

.xo.

.ox.

....

?

o...

.ox.

.xxx

xooo

$

Sample Output

#####

(0,1)

題意：給你一個4*4的棋盤下四子棋，己方下x，對方下o，找出一個必勝局面，輸出第一個落子點。

注：極大極小搜尋一般用在博弈問題中，假定兩位玩家都足夠聰明的情況下，求出最終的得分。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
char mp[5][5];
char str[2] = {'o','x'};
bool check_h(int h,int me)//行判斷
{
    if(mp[h][0] == str[me]&&mp[h][1] == mp[h][2]&&mp[h][2] == mp[h][3]&&mp[h][3] == mp[h][0])
        return true;
    return false;
}
bool check_l(int l,int me)//列判斷
{
    if(mp[0][l] == str[me]&&mp[1][l] == mp[2][l]&&mp[2][l] == mp[3][l]&&mp[3][l] == mp[0][l])
        return true;
    return false;
}
bool check_x(int me)//對角判斷
{
    if(mp[0][0] == mp[1][1]&&mp[1][1] == mp[2][2]&&mp[2][2] == mp[3][3]&&mp[0][0] == str[me])   return true;
    if(mp[0][3] == mp[1][2]&&mp[1][2] == mp[2][1]&&mp[2][1] == mp[3][0]&&mp[0][3] == str[me])   return true;
    return false;
}
bool win(int me)
{
    for(int i = 0; i < 4; i++)
        if(check_h(i,me))
        {
            //cout<<"check_h"<<':';
            //cout<<i<<' '<<me<<endl;
            return true;
        }
        else if(check_l(i,me))
        {
            //cout<<"check_l"<<':';
            //cout<<i<<' '<<me<<endl;
            return true;
        }
        else if(check_x(me))
        {
            //cout<<"check_x"<<':';
            //cout<<i<<' '<<me<<endl;
            return true;
        }
    return false;
}

int ans_x,ans_y,flag;//flag記錄的是對應ans進入的深度
//看別人大佬的程式碼有點看不懂，只能這樣強行判斷了Orz.....
int spa()
{
    int res=0;
    for(int i=0; i<4; i++)
        for(int j=0; j<4; j++)
            if(mp[i][j] == '.')res++;
    return res;
}

int dfs(int me,int deep)
{
    if(!spa())  return 0;
    int minn = 1,maxn = -1;
    for(int i = 0; i < 4; i++)
    for(int j = 0; j < 4; j++)
    {
        if(mp[i][j] == '.')
        {
            mp[i][j] = str[me];
            if(win(me))
            {
                if(flag > deep)
                {
                    ans_x = i;
                    ans_y = j;
                    flag = deep;
                }
                if(flag == deep&&ans_x*3+ans_y > i*3+j)//深度相同時，判斷哪個點更先
                {
                    ans_x = i;
                    ans_y = j;
                }
                mp[i][j] = '.';
                return me == 1?max(maxn,1):min(minn,-1);
            }
            if(me)  maxn = max(maxn,dfs(me^1,deep+1));
            else    minn = min(minn,dfs(me^1,deep+1));
            if(maxn == 1&&flag > deep)
            {
                ans_x = i,ans_y = j;
                flag = deep;
            }
            if(maxn == 1&&flag == deep&&ans_x*3+ans_y > i*3+j)
            {
                ans_x = i,ans_y = j;
            }
            mp[i][j] = '.';
        }
    }
    return me == 1?maxn:minn;
}
int main()
{
    char ch;
    while(scanf(" %c",&ch)&&ch != '$')
    {
        for(int i = 0; i < 4; i++)for(int j = 0; j < 4; j++)scanf(" %c",&mp[i][j]);
        flag = 0x7fffffff;
        if(spa() >= 12) {cout<<"#####"<<endl;continue;}//加了這一行時間瞬間下來了23333
        if(dfs(1,0) == 1)  cout<<'('<<ans_x<<','<<ans_y<<')'<<endl;
        else cout<<"#####"<<endl;
    }
    return 0;
}