FZU2283-Tic-Tac-Toe

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
…
…
…
o
o x o
o . x
x x o
x
o x .
. o .
. . x
o
Sample Output
Cannot win!
Kim win!
Kim win!

题目大意：三子棋，kim先下，问三步之内能否获胜。
解题思路： dfs搜索，参考了斌神的代码，简洁清晰~
dfs(er,step)代表现在er下，若赢返回er，若平局返回0，否则返回3−er（另一个人）赢，注意三者的先后关系。

#include<iostream>#include<cstdio>#include<cmath>#include<vector>#include<map>#include<algorithm>using namespace std;const int MAXN=1e5+5;int mat[5][5];int check(){    for(int i=1;i<=3;i++)    if(mat[i][1]&&mat[i][1]==mat[i][2]&&mat[i][2]==mat[i][3]) return mat[i][1];    for(int j=1;j<=3;j++)    if(mat[1][j]&&mat[1][j]==mat[2][j]&&mat[2][j]==mat[3][j]) return mat[1][j];    if(mat[1][1]&&mat[1][1]==mat[2][2]&&mat[2][2]==mat[3][3]) return mat[1][1];    if(mat[1][3]&&mat[1][3]==mat[2][2]&&mat[2][2]==mat[3][1]) return mat[1][3];    return 0;}int dfs(int er,int step){    int tmp=check();if(tmp!=0) return tmp;    if(step==4) return 0;    bool tie=false;    for(int i=1;i<=3;i++)    {        for(int j=1;j<=3;j++)        {            if(mat[i][j]==0)            {                mat[i][j]=er;                int ans=dfs(3-er,step+1);                mat[i][j]=0;                if(ans==er) return er;                if(ans==0)  tie=true;            }        }    }    if(tie) return 0;    return 3-er;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        char ch[3];        for(int i=1;i<=3;i++)        {            for(int j=1;j<=3;j++)            {                scanf("%s",&ch);                if(ch[0]=='.') mat[i][j]=0;                else if(ch[0]=='o') mat[i][j]=1;                else mat[i][j]=2;            }        }        char op[3];        scanf("%s",op);        int er;        if(op[0]=='o') er=1; else er=2;        if(dfs(er,1)==er) printf("Kim win!\n");        else printf("Cannot win!\n");    }    return 0;}/*3. . .. . .. . .oo x oo . xx x oxo x .. o .. . xo*/