Tic-Tac-Toe（三子连）（总结规律）

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! 

题目：三子连的规则，然后给出残局，

1.kim可以先下一步，判断是否赢，

2.kim下一步，然后对手下一步，kim再下一步 判断是否赢。

刚开始的时候想用DFS因为最近搜索做的多了，然后样例过，WA了3发，发现我只是想当然的算法，然后想到可以分为2种情况 

1.下一步直接获胜

2.”套路“  

*     *     * 

*     *     *

*     *     * 

kim抢了四个角中的一个或者以上+中间的点，然后其余3个角的点再有2个即可套路对方用2步获胜

否则无法判断

#include 
#include      
using namespace std;  
char s[10],juzi[10],ch,ck,m1;        
int judge1()
{
    if(s[0]==ch&&s[1]==ch&&s[2]==ch) return 1;
    if(s[3]==ch&&s[4]==ch&&s[5]==ch) return 1;
    if(s[6]==ch&&s[7]==ch&&s[8]==ch) return 1;
    if(s[0]==ch&&s[3]==ch&&s[6]==ch) return 1;
    if(s[1]==ch&&s[4]==ch&&s[7]==ch) return 1;
    if(s[2]==ch&&s[5]==ch&&s[8]==ch) return 1;
    if(s[0]==ch&&s[4]==ch&&s[8]==ch) return 1;
    if(s[2]==ch&&s[4]==ch&&s[6]==ch) return 1;
    return 0;
}
int judge2()
{   /*0 1 2*/
    /*3 4 5*/
    /*6 7 8*/
    if(s[0]==ch&&s[4]==ch)
    {
        int num=0;
       if(s[2]!=ck)
            num++;
       if(s[6]!=ck)
            num++;
       if(s[8]!=ck)
            num++;
       if(num>1)
        return 1;
    } if(s[6]==ch&&s[4]==ch)
    {
        int num=0;
       if(s[0]!=ck)
            num++;
       if(s[2]!=ck)
            num++;
       if(s[8]!=ck)
            num++;
       if(num>1)
        return 1;
    }
    if(s[2]==ch&&s[4]==ch)
    {
        int num=0;
       if(s[0]!=ck)
            num++;
       if(s[6]!=ck)
            num++;
       if(s[8]!=ck)
            num++;
       if(num>1)
        return 1;
    }
     if(s[4]==ch&&s[8]==ch)
    {
        int num=0;
       if(s[0]!=ck)
            num++;
       if(s[6]!=ck)
            num++;
       if(s[2]!=ck)
            num++;
       if(num>1)
        return 1;
    }
    return 0;

}
int main()
{
   int icase;
   scanf("%d",&icase);
   while(icase--)
   {
       getchar();
       int pos[10],cnt=1,ans=0;
       for(int i=1;i<=9;i++)
       {
          char a;
          scanf("%c",&a);
          getchar();
          s[ans++]=a;
       }
       //¶ÁÈë×Ö·û

      // printf("%s\n",s);

        scanf("%c",&ch);
       if(ch=='x') ck='o';
            else   ck='x';
      for(int i=0;i<9;i++){
          if(s[i]=='.')
          {
              pos[cnt++]=i;
             // printf("%d\n",cnt);
          }

      }  //ÕÒ³ö¿ÕÏÐλÖÃ
      int flag=0;
      for(int i=1;i

这个题目属于数学逻辑推理部分的，是我的弱项，刚看上很是蒙蔽，而且之前有一道三子连的原题在湖科大OJ上没有看过，所以很是紧张，幸好最后AC了。



仍可做的题目：Problem K Wand   Problem G  YYS