Tic-Tac-Toe (eighth session of Fujian Provincial Competition)

problem 2283 Tic-Tac-Toe Accept:2 Submit:3
Time limit:1000 mSec Memory limit:262144 KB problem Description

Kim likes to play tic-tac-toe.

Given A current state, and now Kim was going to take his next move. Please tell Kim if he can win the game in next 2 moves if both player is 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 both players, X and O, who take Turns marking the spaces in a 3x3 grid. The player 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 is 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's a blank place.

Next Line A string ("O" or "x") means Kim was ("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.. .... oo x oo. XX x oxo x.. O.. . XO Sample Output cannot win! Kim win! Kim win! Source

Eighth Fujian University College student Program Design Competition-Replay (thanks to the contractor Xiamen Polytechnic Institute)

At first feel too much trouble, so did not write this problem, the results after the game, review, found that is a water problem, violent search is good, only 3*3 lattice;

Discover or lack of experience, water problems can not be found.

#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <queue
> Using namespace std;
Char map[4][4];
int sum;
int flag;
int ans;
Char A, B; int isOK (char c) {for (int i=0; i<3; i++) {if (map[0][i]==c&&map[1][i]==c&&map[2][i]==c) re       Turn 1;
    Column if (map[i][0]==c&&map[i][1]==c&&map[i][2]==c) return 1;
    } if (map[0][0]==c&&map[1][1]==c&&map[2][2]==c) return 1;           if (map[2][0]==c&&map[1][1]==c&&map[0][2]==c) return 1;
Diagonal return 0;
    } void Dfs () {ans=0;
            for (int i=0, i<3; i++) {for (int j=0; j<3; J + +) {if (map[i][j]!= '. ') continue;
            Map[i][j]=a;
            if (isOK (a)) ans++;
            Map[i][j]= '. ';          if (ans>=2) return;
    There are more than two places for a to win, then a must win}} int main () {int T;
    scanf ("%d", &t);
        while (t--) {flag=0; for (int i=0;i<3;
            i++) {for (int j=0; j<3; J + +) {scanf ("%s", &map[i][j]);
        } getchar ();
        } scanf ("%c", &a);
        if (a== ' o ') b= ' x ';
        else b= ' o ';
            The situation one, already can win at the beginning; if (isOK (a)) {printf ("Kim win!\n");
        Continue }//Case two: Only one can win for (int i=0; i<3; i++) {for (int j=0; j<3; J + +) {if (MA p[i][j]!= '. ')
                Continue
                Map[i][j]=a;
                    if (isOK (a)) {flag=1;
                Break
            } map[i][j]= '. ';
        } if (flag) break;
            } if (flag) {printf ("Kim win!\n");
        Continue
        }//Case three; b only one step can win, so to block this position, and then judge whether there are more than two positions to win, there is a chance to win; ans=0;
                for (int i=0, i<3; i++) for (int j=0; j<3; J + +) {if (map[i][j]!= '. ') continue; MapI
                [J]=b;
                    if (isOK (b)) {flag=1;
                    Map[i][j]=a;
                    DFS ();
                    if (ans>=2) flag=2;

                Map[i][j]= '. ';

            } map[i][j]= '. ';
            } if (flag) {if (flag==2) {printf ("Kim win!\n");
            } else printf ("Cannot win!\n");

        Continue
                }//Is not the above special case, the direct search, see if there can be practiced into line; for (int i=0; i<3; i++) {for (int j=0; j<3; J + +) {
                ans=0;
                    if (map[i][j]!= '. ')
                Continue
                    else {map[i][j]=a;
                    DFS ();
                Map[i][j]= '. ';
            } if (ans>=2) break;
        } if (ans>=2) break;
        } if (ans>=2) printf ("Kim win!\n");
    else printf ("Cannot win!\n");
} return 0;
 }