本文主要是介绍FZU - 2283 (暴力判断)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
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 oSample Output
Cannot win! Kim win! Kim win!
为了不wa，改了好久，才敢提交，写的代码太长了。不知道不修改那些东西，能不能ac。
思路：
幸亏平常喜欢下五子棋，各种赢的套路早就知道了，哈哈。
找出规律后判断就行了。因为只有两步，模拟一下就行了。
1 当在 . 的位置放上棋子后，如果有连续棋子大于等于2的就赢了。
2 如果在 . 的位置放上棋子后，有2个以上的连续位置大于1，那也赢了。（ps：五子棋，经常用这一招绝杀，虽然简单，百试不爽）。
3 其他都是输。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
char map[5][5];
char k;
int pan(int x,int y,char q)
{char q1;if(q=='o')q1='x';else q1='o';int anslie=0,anshang=0,ans1=0,ans2=0;for(int i=0;i<3;i++){if(map[x][i]==q)anslie++;if(map[x][i]==q1)anslie=0;}for(int i=0;i<3;i++){if(map[i][y]==q)anshang++;if(map[i][y]==q1)anshang=0;}if(x==y){if(map[1][1]==q)ans1++;else if(map[1][1]==q1)ans1--;if(map[2][2]==q)ans1++;else if(map[2][2]==q1)ans1--;if(map[0][0]==q)ans1++;else if(map[0][0]==q1)ans1--;}if(x+y==2){if(map[0][2]==q)ans2++;else if(map[0][2]==q1)ans2--;if(map[1][1]==q)ans2++;else if(map[1][1]==q1)ans2--;if(map[2][0]==q)ans2++;else if(map[2][0]==q1)ans2--;}if(anslie>=2||anshang>=2||ans1>=2||ans2>=2){return 1;}int sum=0;if(anslie>=1)sum++;if(anshang>=1)sum++;if(ans1>=1)sum++;if(ans2>=1)sum++;if(sum>=2)return 1;return 0;
}
int work1(char q)
{for(int i=0;i<3;i++){for(int j=0;j<3;j++){if(map[i][j]=='.'){if(pan(i,j,q)){return 1;}}}}return 0;
}
int main()
{int t;scanf("%d",&t);while(t--){for(int i=0;i<3;i++){for(int j=0;j<3;j++){//scanf("%c",&map[i][j]);cin>>map[i][j];}}cin>>k;/*for(int i=0;i<3;i++){printf("%s\n",map[i]);}*/int tem1=work1(k);if(tem1){printf("Kim win!\n");}else{printf("Cannot win!\n");}}return 0;
}这篇关于FZU - 2283 (暴力判断)的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
