Tic-Tac-Toe FZU - 2283

Kim likes to play Tic-Tac-Toe.

Given a current state, and now Kim is going to take his next move. Please tell Kim if he can win the game in next 2 moves if both player are clever enough.

Here “next 2 moves” means Kim’s 2 move. (Kim move,opponent move, Kim move, stop).

Game rules:

Tic-tac-toe (also known as noughts and crosses or Xs and Os) is a paper-and-pencil game for two players, X and O, who take turns marking the spaces in a 3×3 grid. The player who succeeds in placing three of their marks in a horizontal, vertical, or diagonal row wins the game.

Input

First line contains an integer T (1 ≤ T ≤ 10), represents there are T test cases.

For each test case: Each test case contains three lines, each line three string(“o” or “x” or “.”)(All lower case letters.)

x means here is a x

o means here is a o

. means here is a blank place.

Next line a string (“o” or “x”) means Kim is (“o” or “x”) and he is going to take his next move.

Output

For each test case:

If Kim can win in 2 steps, output “Kim win!”

Otherwise output “Cannot win!”

Sample Input

3
. . .
. . .
. . .
o
o x o
o . x
x x o
x
o x .
. o .
. . x
o

Sample Output

Cannot win!
Kim win!
Kim win!

題意就是在走兩步能不能贏 自己模擬不出來 比着隊友的寫的QAQ

#include <iostream>
#include<cstdio>
#include<stack>
#include<cstring>
#include<algorithm>
using namespace std;
char maps[5][5];
int check(char a)
{
    if(maps[0][0]==maps[0][1]&&maps[0][2]==maps[0][0]&&maps[0][0]==a)
        return 1;
    else if(maps[0][0]==maps[1][1]&&maps[2][2]==maps[0][0]&&maps[0][0]==a)
        return 1;
    else   if(maps[1][0]==maps[2][0]&&maps[0][0]==maps[1][0]&&maps[1][0]==a)
        return 1;
    else  if(maps[1][0]==maps[1][1]&&maps[1][2]==maps[1][0]&&maps[1][0]==a)
        return 1;
    else  if(maps[2][0]==maps[2][1]&&maps[2][0]==maps[2][2]&&maps[2][0]==a)
        return 1;
    else if(maps[0][1]==maps[1][1]&&maps[2][1]==maps[0][1]&&maps[0][1]==a)
        return 1;
    else  if(maps[0][2]==maps[1][2]&&maps[2][2]==maps[0][2]&&maps[0][2]==a)
        return 1;
    else  if(maps[0][2]==maps[1][1]&&maps[2][0]==maps[0][2]&&maps[0][2]==a)
        return 1;
    else
        return 0;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        for(int i=0;i<3;i++)
        {
            for(int j=0;j<3;j++)
            cin>>maps[i][j];
        }
        char a;
        cin>>a;
        int count=0,flag1=0,flag2=0;
        for(int i=0;i<3;i++)
        {
            for(int j=0;j<3;j++)
            {
                if(maps[i][j]=='.')  //下第一步 這時隊手沒下 所以如果可以練成一條線就贏了
                {


                    maps[i][j]=a;              //和開場贏一起考慮在內了
                    if(check(a))
                    {
                        flag1=1;
                        break;
                    }

                count=0;   //用於數有幾種贏法
                for(int k=0;k<3;k++)
                {
                    for(int l=0;l<3;l++)
                    {
                        if(maps[k][l]=='.')
                        {
                            maps[k][l]=a;
                            if(check(a))
                            {
                                count++;
                            }
                            if(count>1)   //贏法有兩種就能贏
                            {
                                flag2=1;
                                break;
                            }
                            maps[k][l]='.'; // 回溯
                        }

                    }
                    if(flag2)break;
                }
                maps[i][j]='.';  //第一步回溯
                }
            }
            if(flag1||flag2)break;
        }
        if(flag1||flag2)
            cout<<"Kim win!"<<endl;
        else
            cout<<"Cannot win!"<<endl;

    }
    return 0;
}