【FZUoj 2283 Tic-Tac-Toe】& dfs & 博弈

Problem 2283 Tic-Tac-Toe
Accept: 159 Submit: 340
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!
Source

第八届福建省大学生程序设计竞赛-重现赛（感谢承办方厦门理工学院）

题意 ： 给出一个 3 * 3 的 棋局，以及先手的棋子 o ,轮到先手下，有三枚棋子连着者获胜，问先手下次，对方下一次，先手再下一次，先手能否获胜

思路 ： 先手已经有两枚棋子连着了且第三个点没有对方的落子，是否存在一个 ‘.’，先手在该位置落子后可以形成至少在两条线上有两枚自己的落子（算上刚刚下）,以上情况可获胜，其余不能获胜（给出的棋局似乎没有已经有三枚相同的棋子相连的情况，似乎也没有对方 已有两枚举连着的棋子，先手下一次必须去堵一下情况）

AC代码：

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;const int MAX = 1e5 + 10;typedef long long LL;char s[4][4],c[2],ss[2];int ok,fx[20]= {-1,1,-1,1,0,0,1,2,0,0,1,2,-1,-2,0,0,-1,-2},fy[20] = {-1,1,0,0,-1,1,1,2,1,2,0,0,-1,-2,-1,-2,0,0};void dfs(int x,int y,char o,int sum){   int p = 0;   for(int i = 0; i < 18; i += 2){        int ans = 0,xx = x + fx[i],yy = y + fy[i],x2 = x + fx[i + 1],y2 = y + fy[i + 1];        if(xx >= 0 && yy >= 0 && xx < 3 && yy < 3 && x2 >= 0 && y2 >= 0 && x2 < 3 && y2 < 3){            if(s[xx][yy] != o && s[xx][yy] != '.') continue;            if(s[x2][y2] != o && s[x2][y2] != '.') continue;            if(s[xx][yy] == o) ans++;            if(s[x2][y2] == o) ans++;            if(ans) p++;            if(ans + sum == 2 || p == 2){ ok = 1; return ;}         }   }}int main(){    int T; scanf("%d",&T);    while(T--){        for(int i = 0; i < 3; i++)            for(int j = 0; j < 3 ;j++)                scanf("%s",ss),s[i][j] = ss[0];        scanf("%s",c);        ok = 0;        for(int i = 0; i < 3 && !ok; i++)            for(int j = 0; j < 3 && !ok; j++)                if(s[i][j] == c[0] || s[i][j] == '.')                    if(s[i][j] == c[0]) dfs(i,j,c[0],1);                    else dfs(i,j,c[0],0);        if(ok) puts("Kim win!");        else puts("Cannot win!");    }    return 0;}