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.
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.
For each test case:
If Kim can win in 2 steps, output “Kim win!”
Otherwise output “Cannot win!”
3 . . . . . . . . . o o x o o . x x x o x o x . . o . . . x oSample Output
Cannot win! Kim win!
Kim win!
题意：输入一个棋盘，判断两步之内能不能赢，每行每列或者对角线有三个子就算赢
为数不多的没看题解就自己写出来的题，有点小激动诶！！！咳咳......下面说下我的思路：
其实说是判断两步之内能不能赢，但是只要下一步就能判断出来能不能赢，举个栗子说：下五子棋的时候，如果有四个子连成一条线，并且两边上没有阻碍的话，这局棋胜负已经决定了，就无需再下了
就是说如果只有一个位置下完之后两个子，对方不想让你赢，肯定会去堵你，这样你就赢不了，但是如果有两个位置已经是两个子了，无论对面堵那个，你都会赢
例：x . x 无论怎么下都是死棋了，话不多说，上代码！！！
. . .
x . .
#include<iostream>
using namespace std;
char a[4][4];
char s;
int dfs(int x,int y,char w) //并不是dfs，只是顺手起的名
{
int mark=0; //标记有几个点快要完成
for(int i=1;i<=3;i++) //判断每行是否有符合要求的
{
int w1=0,w1s=0;
for(int j=1;j<=3;j++)
{
if(a[i][j]==w) w1++;
else if(a[i][j]=='.') w1s++;
if(w1==3) return 1; //如果已经有三个点直接退出
else if(w1==2&&w1s==1) mark++; //两个点和一个'.'符合要求
}
}
for(int i=1;i<=3;i++) //判断每列是否有符合要求的
{
int w2=0,w2s=0;
for(int j=1;j<=3;j++)
{
if(a[j][i]==w) w2++;
else if(a[j][i]=='.') w2s++;
if(w2==3) return 1;
else if(w2==2&&w2s==1) mark++;
}
}
for(int i=1;i<=3;i++) //判断主对角线
{
int w3=0,w3s=0;
if(a[i][i]==w) w3++;
else if(a[i][i]=='.') w3s++;
if(w3==3) return 1;
else if(w3==2&&w3s==1) mark++;
}
for(int i=1;i<=3;i++) //判断副对角线
{
int w4=0,w4s=0;
if(a[i][4-i]==w) w4++;
else if(a[i][4-i]=='.') w4s++;
if(w4==3) return 1;
else if(w4==2&&w4s==1) mark++;
}
if(mark>=2) //超过两处直接退出
return 1;
return 0;
}
int main()
{
int t;
while(cin>>t)
{
while(t--)
{
for(int i=1;i<=3;i++)
{
for(int j=1;j<=3;j++)
{
cin>>a[i][j];
}
}
cin>>s;
int ok;
for(int i=1;i<=3;i++)
{
for(int j=1;j<=3;j++)
{
ok=0;
if(a[i][j]=='.')
{
a[i][j]=s; //假设将棋下在此处
if(dfs(i,j,s)==1)
{
ok=1;
break;
}
a[i][j]='.'; //如果下在此处不符合要求，将标记删除！！！
}
}
if(ok==1)
break;
}
if(ok==1)
cout<<"Kim win!"<<endl;
else
cout<<"Cannot win!"<<endl;
}
}
}
/**
_ooOoo_
o8888888o
88" . "88
(| -_- |)
O\ = /O
____/`---'\____
.' \\| |// `.
/ \\||| : |||// \
/ _||||| -:- |||||- \
| | \\\ - /// | |
| \_| ''\---/'' | |
\ .-\__ `-` ___/-. /
___`. .' /--.--\ `. . __
."" '< `.___\_<|>_/___.' >'"".
| | : `- \`.;`\ _ /`;.`/ - ` : | |
\ \ `-. \_ __\ /__ _/ .-` / /
======`-.____`-.___\_____/___.-`____.-'======
`=---='
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
佛祖保佑 每次AC
**/