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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.
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.
For each test case:
If Kim can win in 2 steps, output “Kim win!”
Otherwise output “Cannot win!”
3 . . . . . . . . . o o x o o . x x x o x o x . . o . . . x o
Cannot win! Kim win! Kim win!
题意就是在走两步能不能赢 自己模拟不出来 比着队友的写的QAQ#include <iostream> #include<cstdio> #include<stack> #include<cstring> #include<algorithm> using namespace std; char maps[5][5]; int check(char a) {if(maps[0][0]==maps[0][1]&&maps[0][2]==maps[0][0]&&maps[0][0]==a)return 1;else if(maps[0][0]==maps[1][1]&&maps[2][2]==maps[0][0]&&maps[0][0]==a)return 1;else if(maps[1][0]==maps[2][0]&&maps[0][0]==maps[1][0]&&maps[1][0]==a)return 1;else if(maps[1][0]==maps[1][1]&&maps[1][2]==maps[1][0]&&maps[1][0]==a)return 1;else if(maps[2][0]==maps[2][1]&&maps[2][0]==maps[2][2]&&maps[2][0]==a)return 1;else if(maps[0][1]==maps[1][1]&&maps[2][1]==maps[0][1]&&maps[0][1]==a)return 1;else if(maps[0][2]==maps[1][2]&&maps[2][2]==maps[0][2]&&maps[0][2]==a)return 1;else if(maps[0][2]==maps[1][1]&&maps[2][0]==maps[0][2]&&maps[0][2]==a)return 1;elsereturn 0; } int main() {int t;cin>>t;while(t--){for(int i=0;i<3;i++){for(int j=0;j<3;j++)cin>>maps[i][j];}char a;cin>>a;int count=0,flag1=0,flag2=0;for(int i=0;i<3;i++){for(int j=0;j<3;j++){if(maps[i][j]=='.') //下第一步 这时队手没下 所以如果可以练成一条线就赢了{maps[i][j]=a; //和开场赢一起考虑在内了if(check(a)){flag1=1;break;}count=0; //用于数有几种赢法for(int k=0;k<3;k++){for(int l=0;l<3;l++){if(maps[k][l]=='.'){maps[k][l]=a;if(check(a)){count++;}if(count>1) //赢法有两种就能赢{flag2=1;break;}maps[k][l]='.'; // 回溯}}if(flag2)break;}maps[i][j]='.'; //第一步回溯}}if(flag1||flag2)break;}if(flag1||flag2)cout<<"Kim win!"<<endl;elsecout<<"Cannot win!"<<endl;}return 0; }
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