Tic-Tac-Toe (第八届福建省赛)

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 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. . .. . .. . .oo x oo . xx x oxo x .. o .. . xo

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）  

一开始觉得太麻烦了，所以没去写这道题，结果比赛完后，回顾了一下，发现就是一道水题，暴力搜索就好了，只有3*3的格子；

发现还是缺乏经验，水题都发现不了。

#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) return 1;       //行列        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;           //对角线    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;          //有超过两个地方可以让A赢，则A必赢        }    }}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';        //情况一；一开始便已经可以赢了；        if(isok(a)) {            printf("Kim win!\n");            continue;        }        //情况二：只差一位就可以赢        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)) {                    flag=1;                    break;                }        map[i][j]='.';            }            if(flag)                break;        }        if(flag) {            printf("Kim win!\n");            continue;        }        // 情况三；b只差一步就可以赢，所以要封锁这个位置，然后判断之后是否有超过两个位置可以赢，有则可以赢；        ans=0;        for(int i=0; i<3; i++)            for(int j=0; j<3; j++) {                if(map[i][j]!='.') continue;                map[i][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;        }          //不是以上特殊情况，则直接搜索，看有没有能练成线的；        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;}