FZU Tic-Tac-Toe

Problem L Tic-Tac-Toe

Accept: 94    Submit: 184
Time Limit: 1000 mSec    Memory Limit : 262144 KB

 Problem Description

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!

代码有点糙，爆搜过，思路如下：

.为0，x为-1，o为1

设一个系数k，当我方为x的时候k为-1，为o时k为1.

判断行列斜和，若和为2直接true，和为1用cnt计数下来，当cnt大于等于2时候（其实2就可以了）也为true。

代码如下：

#include <iostream>#include <stdio.h>#include <algorithm>#include <string.h>#include <cmath>using namespace std;char map[4][4];int s[4][4];int t;int sum;int k;int cnt;bool judge(int x,int y){    if((x+y)%2==1){        cnt=0,sum=0;        for(int i=1;i<=3;i++){            sum+=s[i][y];        }        sum*=k;        if(sum==2)            return true;        else if(sum==1)            cnt++;        sum=0;        for(int i=1;i<=3;i++){            sum+=s[x][i];        }        sum*=k;        if(sum==2)            return true;        else if(sum==1)            cnt++;        if(cnt==2){            return true;        }    }    else {        cnt=0,sum=0;        for(int i=1;i<=3;i++){            sum+=s[i][y];        }        sum*=k;        if(sum==2)            return true;        else if(sum==1)            cnt++;        sum=0;        for(int i=1;i<=3;i++){            sum+=s[x][i];        }        sum*=k;        if(sum==2)            return true;        else if(sum==1)            cnt++;        sum=0;        if(x==y){            for(int i=1;i<=3;i++){                sum+=s[i][i];            }            sum*=k;                        if(sum==2)                return true;            else if(sum==1)                cnt++;        }        else {            for(int i=1;i<=3;i++){                sum+=s[i][4-i];            }            sum*=k;                        if(sum==2)                return true;            else if(sum==1)                cnt++;        }        if(cnt>=2){            return true;        }    }    return false;}int main(){    char st;    int flag;    scanf("%d",&t);    while(t--){        flag=0;        memset(s,0,sizeof(0));        for(int i=1;i<=3;i++){            for(int j=1;j<=3;j++){                cin>>map[i][j];                if(map[i][j]=='.')  s[i][j]=0;                else if(map[i][j]=='o') s[i][j]=1;                else if(map[i][j]=='x') s[i][j]=-1;            }        }        cin>>st;        if(st=='o') k=1;        else if (st=='x') k=-1;        for(int i=1;i<=3;i++){            for(int j=1;j<=3;j++){                if(map[i][j]=='.'){                    if(judge(i,j))                        flag=1;                }            }        }        if(flag) puts("Kim win!");            else    puts("Cannot win!");    }    return 0;}