FZU Tic-Tac-Toe -.- FZU邀请赛 FZU 2283

最新推荐文章于 2018-05-01 19:57:36 发布

转载最新推荐文章于 2018-05-01 19:57:36 发布 · 66 阅读

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原文链接：http://www.cnblogs.com/l609929321/p/7296104.html

本文介绍了一种判断九宫棋（Tic-Tac-Toe）中玩家Kim是否能在接下来的两步内获胜的方法。通过分析棋盘状态和玩家下一步行动，算法能够预测玩家是否有机会赢得比赛。

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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;  
}

转载于:https://www.cnblogs.com/l609929321/p/7296104.html