HDU-1312 Red and Black

原创已于 2022-11-18 09:56:38 修改 · 125 阅读

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#算法 #数据结构 #bfs #广度搜索 #c++

于 2021-03-18 09:25:28 首次发布

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本文介绍了一道经典的迷宫寻路问题HDU-1312，利用宽度优先搜索(BFS)算法计算从起始位置可达的所有黑色方块数量。输入包含多个测试案例，每个案例由地图尺寸及地图本身组成，地图中用字符表示不同类型的方块，最终输出可达黑色方块总数。

HDU-1312 Red and Black

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can’t move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above.

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

‘.’ - a black tile
‘#’ - a red tile
‘@’ - a man on a black tile(appears exactly once in a data set)

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

Sample Input

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

Sample Output

my analysis

This problem is limited to memory, so we can’t use the conventional tag array. After each visit, we can directly reset this block as an obstacle, and use BFS once to connect all the accessible points

solution（C++）

#include<iostream>
#include<queue>
using namespace std;
int A, B, sum = 1;
char maze[10000][10000];
int X[4] = { 0,0,1,-1 };
int Y[4] = { 1,-1,0,0 };
struct node
{
	int x;
	int y;
}Node, S;
bool judge(int x, int y)
{
	if (x < 0 || x >= A || y < 0 || y >= B || maze[x][y] == '#' || maze[x][y] == '@')
		return false;
	else
		return true;
}
void BFS()
{
	queue<node> q;
	q.push(S);
	while (!q.empty())
	{
		node top = q.front();
		q.pop();
		for (int i = 0; i < 4; i++)
		{
			int newX = top.x + X[i];
			int newY = top.y + Y[i];
			if (judge(newX, newY))
			{
				sum++;
				Node.x = newX, Node.y = newY;
				q.push(Node);
				maze[newX][newY] = '#';
			}
		}
	}
	return;
}
int main()
{
	while (cin >> B >> A)
	{
		if (A == 0 && B == 0)
			break;

		sum = 1;
		for (int i = 0; i < A; i++)
			for (int j = 0; j < B; j++)
			{
				cin >> maze[i][j];
				if (maze[i][j] == '@')
				{
					S.x = i;
					S.y = j;
				}
			}
		BFS();
		cout << sum << endl;
	}
}