CF - 754B. Ilya and tic-tac-toe game 模拟+暴力

最新推荐文章于 2020-04-04 11:58:11 发布

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最新推荐文章于 2020-04-04 11:58:11 发布

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分类专栏：模拟暴力文章标签： acm codeforces 算法

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本文介绍了一个简单的4x4井字棋游戏，分析了如何判断玩家Ilya是否能在当前局面通过一步操作赢得比赛。通过遍历所有空位并尝试放置Ilya的标记X，文章提供了一种有效的方法来确定是否存在获胜的可能性。

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1.题目描述：

Ilya is an experienced player in tic-tac-toe on the 4 × 4 field. He always starts and plays with Xs. He played a lot of games today with his friend Arseny. The friends became tired and didn't finish the last game. It was Ilya's turn in the game when they left it. Determine whether Ilya could have won the game by making single turn or not.

The rules of tic-tac-toe on the 4 × 4 field are as follows. Before the first turn all the field cells are empty. The two players take turns placing their signs into empty cells (the first player places Xs, the second player places Os). The player who places Xs goes first, the another one goes second. The winner is the player who first gets three of his signs in a row next to each other (horizontal, vertical or diagonal).

Input

The tic-tac-toe position is given in four lines.

Each of these lines contains four characters. Each character is '.' (empty cell), 'x' (lowercase English letter x), or 'o' (lowercase English letter o). It is guaranteed that the position is reachable playing tic-tac-toe, and it is Ilya's turn now (in particular, it means that the game is not finished). It is possible that all the cells are empty, it means that the friends left without making single turn.

Output

Print single line: "YES" in case Ilya could have won by making single turn, and "NO" otherwise.

Examples

input
xx..
.oo.
x...
oox.

output
YES

input
x.ox
ox..
x.o.
oo.x

output
NO

input
x..x
..oo
o...
x.xo

output
YES

input
o.x.
o...
.x..
ooxx

output
NO

Note

In the first example Ilya had two winning moves: to the empty cell in the left column and to the leftmost empty cell in the first row.

In the second example it wasn't possible to win by making single turn.

In the third example Ilya could have won by placing X in the last row between two existing Xs.

In the fourth example it wasn't possible to win by making single turn.

2.题目概述：

类似于五子棋，三连则胜利，然后玩到一半朋友走了，要你判断他（"x"方）是否只下一步就能赢

3.解题思路：

地图很小直接暴力对“.”的位置下x，每下一步就判断是否会胜利即可，注意回溯。判断胜利条件就是看周围八个位置是否连着，应该还可以剪枝？

4.AC代码：

#include <stdio.h>
using namespace std;
char mp[4][4];
int judge()
{
	for (int i = 0; i < 4; i++)	// 横着
		for (int j = 0; j <= 1; j++)
			if (mp[i][j] == 'x' && mp[i][j + 1] == 'x' && mp[i][j + 2] == 'x')
				return 1;

	for (int j = 0; j <= 1; j++) // 竖着
		for (int i = 0; i < 4; i++)
			if (mp[j][i] == 'x' && mp[j + 1][i] == 'x' && mp[j + 2][i] == 'x')
				return 1;

	for (int i = 0; i <= 1; i++)  // 斜着
		for (int j = 0; j <= 1; j++)
			if (mp[i][j] == 'x' && mp[i + 1][j + 1] == 'x' && mp[i + 2][j + 2] == 'x')
				return 1;

	for (int i = 0; i <= 1; i++) // 斜着
		for (int j = 2; j < 4; j++)
			if (mp[i][j] == 'x' && mp[i + 1][j - 1] == 'x' && mp[i + 2][j - 2] == 'x')
				return 1;
	return 0;
}

int main()
{
	int flag = 0;
	for (int i = 0; i < 4; i++)
		scanf("%s", mp[i]);
	for (int i = 0; i < 4; i++)
		for (int j = 0; j < 4; j++)
		{
			if (mp[i][j] == '.')
			{
				mp[i][j] = 'x';
				if (judge())
				{
					flag = 1;
					break;
				}
				mp[i][j] = '.';
			}
		}
	if (flag)
		puts("YES");
	else
		puts("NO");
	return 0;
}