C++实现——三子棋游戏

/*
题目描述：
两个人玩三子棋游戏,即在3*3的矩阵上下棋，一个人画叉一个人画圈，谁先出现成行或成列或成对角线三个相同的棋子就算谁赢。编写算法实现，判断给定棋局的状态，用1代表先手，2代表后手。出现的六种状态为 1won 2won x（代表棋局错误） draw（代表平局） 1（下一步先手走） 2（下一步后手走）

*/

输入 ：(共含有三种字符 x . 0)
含有测试，每组测试用例需要输入三行
样例：
x.x
x0x
0x.
输出：
2

#include "stdafx.h"

#include 
#include 
#include 
#include
#include 
#include 
using namespace std;

int main(){
    //代表行矩阵
    vector<vector<char>> board1(3, vector<char>(3));
    //代表列矩阵
    vector<vector<char>> board2(3, vector<char>(3));

    //记录主/副对角线
    vector<char> a, b;
    while (1){
        int onum = 0;
        int xnum = 0;
        for (int i = 0; i < 3; i++){

            for (int j = 0; j < 3; j++){
                char c;
                cin >> c;
                //填充行矩阵
                board1[i][j] = c;
                //填充列矩阵
                board2[j][i] = c;
                //统计先手和后手放棋子的个数
                if (board1[i][j] == '0')++onum;
                if (board1[i][j] == 'x')++xnum;

                //记录正对角线
                if (i == j)a.push_back(c);
                //记录斜对角线
                if ((i == 1 && j == 1) || (i == 0 && j == 2) || (i == 2 && j == 0))
                    b.push_back(c);

            }
        }
        vector<char> t1 = { '0', '0', '0' }, t2 = {'x','x','x'};
        int count = 0;
        for (int i = 0; i < 3; i++){

            if (board1[i] == t1 || board1[i] == t2) ++count;
            if (board2[i] == t1 || board2[i] == t2) ++count;
        }
        if (t1 == a || t2==a)++count;
        if (t1 == b || t2 == b)++count;
        //不合理的棋局
        if (count >= 2){

            cout << "x" << endl;
        }
        //由一方剩
        else if(count==1){
            if (onum == xnum)cout << "2 won" << endl;
            else cout << "1 won" << endl;

        }
        //平局或者继续走
        else {
            //棋盘已满，不分胜负
            if (onum + xnum == 9)cout << "draw" << endl;
            else{
                 //恰为偶数，轮着1走
                if (onum == xnum)cout << "1" << endl;
                //为奇数，轮着2走
                else cout << "2" << endl;
            }
        }


    }

    return 0;
}