FZU Tic-Tac-Toe -.- FZU邀请赛 FZU 2283

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

. . .

o x o

o . x

x x o

o x .

. o .

. . x

Sample Output

Cannot win!

Kim win!

九宫棋Kim先下两步之内是否可以胜利

题解下次贴

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <cmath>
using namespace std;
char map[4][4];
int s[4][4],t,sum,k,cnt;
bool judge(int x,int y){
    if((x+y)%2==1){//当此时这个格子的行列的和为奇数时
        cnt=0,sum=0; //cnt代表可能胜利的次数
        for(int i=1;i<=3;i++){
            sum+=s[i][y];
        }
        sum*=k;  //同是负数相乘为正数
        if(sum==2)  //空白数为2时代表可以
            return true;
        else if(sum==1)  //当sum和为1时，此时可能胜利
            cnt++;
        sum=0;  //注意，此时一定要重置sum，因为sum在这个函数的前面后面的含义不同
        for(int i=1;i<=3;i++){
            sum+=s[x][i];//计算此时画相同的符号的个数之和
        }
        sum*=k;
        if(sum==2)  //如果画相同符号的和为2，代表画下一个一定会胜利
            return true;
        else if(sum==1)  //如果此时画的相同的符合为1，代表可能胜利
            cnt++;
        if(cnt==2){  //当可能胜利的次数超过2时一定可以胜利
            return true;
        }
    }
    else {  //当此时这个各自的行列的和为偶数时
        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){  //代表这个空白所在的地方在斜线上
            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;  //代表此时Kim所用的符号
        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;
}