【FZU

L - Tic-Tac-Toe

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. . .. . .. . .oo x oo . xx x oxo x .. o .. . xo

Sample Output

Cannot win!Kim win!Kim win!

题意：给出一个3x3的棋盘，两人用'x'和'o'下棋，三个棋子连成一条线就算赢，现在给出Kim的棋子样式，问Kim能否在两步之内取得胜利（若没有一步取胜，对手也会再下一步）。

分析：我们可以遍历棋盘上的点，若点为'.'则将其变为Kim的棋子进行一次判断，是否存在三子连珠，若存在则说明Kim可以赢得比赛。如果改变点后未形成三子连珠，则再次进行遍历，将'.'变为Kim的棋子，判断是否有两个以上的点符合三子连珠，若存在则Kim可以赢得比赛。

因为一步之后对手会下一步棋，如果只存在一种满足的情况，那么对手会将路堵死，所以要有两点以上满足条件。

代码如下：

#include <map>#include <cmath>#include <queue>#include <stack>#include <vector>#include <cstdio>#include <string>#include <cstring>#include <iostream>#include <algorithm>#define LL long long#define INF 0x3f3f3f3fusing namespace std;const int MX = 5;char mp[MX][MX];char op[5];char read(){    char c = getchar();    if(c == ' ' || c == '\n'){        c = getchar();    }    return c;}int check(){    for(int i = 1; i <= 3; i++){        for(int j = 1; j <= 3; j++){            if(mp[i][j] == op[0]){                if(i+2 <= 3 && mp[i+1][j] == op[0] && mp[i+2][j] == op[0])  return 1;                if(j+2 <= 3 && mp[i][j+1] == op[0] && mp[i][j+2] == op[0])  return 1;                if(i+2 <= 3 && j+2 <= 3 && mp[i+1][j+1] == op[0] && mp[i+2][j+2] == op[0])  return 1;                if(i+2 <= 3 && j-3 >= 0 && mp[i+1][j-i] == op[0] && mp[i+2][j-2] == op[0])  return 1;            }        }    }    return 0;}int main(){    int t;    scanf("%d", &t);    while(t--){        for(int i = 1; i <= 3; i++){            for(int j = 1; j <= 3; j++){                mp[i][j] = read();            }        }        scanf("%s", op);        int flag = 0;        for(int i = 1; i <= 3; i++){            for(int j = 1; j <= 3; j++){                if(mp[i][j] == '.'){                    int cnt = 0;                    mp[i][j] = op[0];                    if(check()){                        flag = 1;                        break;                    }                    else{                        for(int p = 1; p <= 3; p++){                            for(int q = 1; q <= 3; q++){                                if(mp[p][q] == '.'){                                    mp[p][q] = op[0];                                    if(check()){                                        cnt++;                                    }                                    mp[p][q] = '.';                                }                            }                        }                        if(cnt >= 2){                            flag = 1;                            break;                        }                    }                    mp[i][j] = '.';                }            }        }        if(flag)    puts("Kim win!");        else puts("Cannot win!");    }    return 0;}