Tic Tac Toe

2024-02-02 02:48

文章标签 tic tac toe

本文主要是介绍Tic Tac Toe，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

文章目录

一、题目描述
二、题目理解
三、代码展示

一、题目描述

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?
Input
The first line of input contains N, the number of test cases. 4N-1 lines follow, specifying N grid configurations separated by empty lines.
Output
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
Sample Output
yes
no

二、题目理解

在这里插入图片描述

三、代码展示

#include<stdio.h>
char plant[4][4];
int win(char c)
{int i,j;for(i=0;i<3;i++){for(j=0;j<3&&plant[i][j]==c;j++){if(j==2)return 1;}for(j=0;j<3&&plant[j][i]==c;j++){if(j==2)return 1;}}for(i=0;i<3&&plant[i][i]==c;i++){if(i==2)return 1;}for(i=0;i<3&&plant[i][2-i]==2;i++){if(i==2)return 1;}return 0;
}
int main()
{int n,x,o;int flag;int i,j;while(scanf("%d",&n)!=EOF){getchar();while(n--){x=0;o=0;for(i=0;i<3;i++){scanf("%s",plant[i]);}for(i=0;i<3;i++){for(j=0;j<3;j++){if(plant[i][j]=='X'){x++;}if(plant[i][j]=='O'){o++;}}}flag=1;if(win('X')&&win('O')){flag=0;}if(win('X')&&x==o){flag=0;}if(win('O')&&x!=o){flag=0;}if(o>x||x-o>1){flag=0;}if(flag==1)printf("yes\n");if(flag==0)printf("no\n");}}return 0;
}