Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem B: Tic Tac Toe

Tic Tac Toe is a child's game played on a 3 by 3 grid. One player, X, starts by placing an X at an unoccupied grid position. Then the other player, O, places an O at an unoccupied grid position. Play alternates between X and O until the grid is filled or one player's symbols occupy an entire line (vertical, horizontal, or diagonal) in the grid.

We will denote the initial empty Tic Tac Toe grid with nine dots. Whenever X or O plays we fill in an X or an O in the appropriate position. The example below illustrates each grid configuration from the beginning to the end of a game in which X wins.

...  X..  X.O  X.O  X.O  X.O  X.O  X.O
...  ...  ...  ...  .O.  .O.  OO.  OO.
...  ...  ...  ..X  ..X  X.X  X.X  XXX

Your job is to read a grid and to determine whether or not it could possibly be part of a valid Tic Tac Toe game. That is, is there a series of plays that can yield this grid somewhere between the start and end of the game?

The first line of input contains N, the number of test cases. 4N-1 lines follow, specifying N grid configurations separated by empty lines. For each case print "yes" or "no" on a line by itself, indicating whether or not the configuration could be part of a Tic Tac Toe game.

Sample Input

2
X.O
OO.
XXX

O.X
XX.
OOO

Output for Sample Input

yes
no

---

一定由 X 先下，然後換 O。
no 的情況
1. 不可能 X win, O win
2. X win 時 x == o
3. O win 時 x == o+1

#include <stdio.h>
int check(char g[][4]) {
    int i, j, o = 0, x = 0;
    for(i = 0; i < 3; i++)
        for(j = 0; j < 3; j++)
            if(g[i][j] == 'O')
                o++;
            else if(g[i][j] == 'X')
                x++;
    if(!(x == o || x == o+1))
        return 0;
    int oflag = 0, xflag = 0;
    for(i = 0; i < 3; i++) {
        if(g[i][0] == 'O' && g[i][0] == g[i][1] && g[i][1] == g[i][2])
            oflag = 1;
        if(g[0][i] == 'O' && g[0][i] == g[1][i] && g[1][i] == g[2][i])
            oflag = 1;
        if(g[i][0] == 'X' && g[i][0] == g[i][1] && g[i][1] == g[i][2])
            xflag = 1;
        if(g[0][i] == 'X' && g[0][i] == g[1][i] && g[1][i] == g[2][i])
            xflag = 1;
    }
    if(g[0][0] == 'O' && g[0][0] == g[1][1] && g[1][1] == g[2][2])
        oflag = 1;
    if(g[2][0] == 'O' && g[2][0] == g[1][1] && g[1][1] == g[0][2])
        oflag = 1;
    if(g[0][0] == 'X' && g[0][0] == g[1][1] && g[1][1] == g[2][2])
        xflag = 1;
    if(g[2][0] == 'X' && g[2][0] == g[1][1] && g[1][1] == g[0][2])
        xflag = 1;
    if(oflag && xflag)
        return 0;
    if(!oflag && !xflag)
        return 1;
    if(xflag && x == o+1)
        return 1;
    if(oflag && x == o)
        return 1;
    return 0;
}
int main() {
    int t;
    char g[4][4];
    scanf("%d", &t);
    while(t--) {
        scanf("%s%s%s", g[0], g[1], g[2]);
        if(check(g))
            puts("yes");
        else
            puts("no");
    }
    return 0;
}