#27-LETTERS[dfs](zly#4)

最新推荐文章于 2024-03-16 21:55:17 发布

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本文介绍了一款单人棋盘游戏的最优路径搜索算法。游戏在一个R行C列的矩形棋盘上进行，棋盘上的每个位置都有一个大写字母。玩家从左上角开始移动棋子，目标是在不重复进入相同字母标记的位置的前提下，尽可能多地移动。文章通过深度优先搜索算法实现了这一目标，并提供了一个完整的代码实现。

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Description

A single-player game is played on a rectangular board divided in R rows and C columns. There is a single uppercase letter (A-Z) written in every position in the board.
Before the begging of the game there is a figure in the upper-left corner of the board (first row, first column). In every move, a player can move the figure to the one of the adjacent positions (up, down,left or right). Only constraint is that a figure cannot visit a position marked with the same letter twice.
The goal of the game is to play as many moves as possible.
Write a program that will calculate the maximal number of positions in the board the figure can visit in a single game.

Input

The first line of the input contains two integers R and C, separated by a single blank character, 1 <= R, S <= 20.
The following R lines contain S characters each. Each line represents one row in the board.

Output

The first and only line of the output should contain the maximal number of position in the board the figure can visit.

Sample Input

3 6
HFDFFB
AJHGDH
DGAGEH

Sample Output

Code:

#include <iostream>

#define SIZE 25

using namespace std;

int res = 1, n, m;
int dx[4] = {1, -1, 0, 0};
int dy[4] = {0, 0, 1, -1};
char a[SIZE][SIZE];
bool v[128] = {true};

void dfs(int x, int y, int step) // 深搜
{
	int i, r, c;
	
	if (step > res)
	{
		res = step;
	}
	for (i = 0; i < 4; i++)
	{
		r = x + dx[i];
		c = y + dy[i];
		if (!v[a[r][c]])
		{
			v[a[r][c]] = true; // 保存结果
			dfs(r, c, step + 1); // 递归
			v[a[r][c]] = false; // 回溯
		}
	}
}

int main(int argc, char** argv)
{
	int i, j;
	
	cin >> n >> m;
	for (i = 1; i <= n; i++) // 输入
	{
		for (j = 1; j <= m; j++)
		{
			cin >> a[i][j];
		}
	}
	
	v[a[1][1]] = true;
	dfs(1, 1, 1); // 开始深搜
	
	cout << res << endl; // 输出结果
	
	return 0;
}