ACM--dfs与bfs

最新推荐文章于 2022-05-06 00:48:01 发布

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本文介绍了一种使用深度优先搜索（DFS）算法来解决二维网格中寻找最大连通块的问题。连通块由相邻的填充单元格组成，包括水平、垂直和对角线方向的邻居。算法的目标是在给定的网格上找到最大的连通块，即包含最多填充单元格的区域。

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1.dfs求最大连通块（UVa 871 - Counting Cells in a Blob）

Consider a two-dimensional grid of cells, each of which may be empty
or filled. Filled cells form blobs. The filled cells that are connected
form the same bigger blob. Two cells are said to be connected if
they are adjacent to each other horizontally, vertically, or diagonally.
There may be several blobs on the grid. Your job is to find the largest
blob (in terms of number of cells) on the grid.
Write a program that determines the size of the largest blob for
a given set of blobs.
Input
The input begins with a single positive integer on a line by itself
indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also
a blank line between two consecutive inputs.
The grid is given as a set of strings, each composed of 0’s and 1’s. The ‘1’ indicates that the cell is
filled and ‘0’ indicates an empty cell. The strings should be converted into the grid format.
The largest grid that sould be considered is a 25×25 grid.
Output
For each test case, the output must follow the description below. The outputs of two consecutive cases
will be separated by a blank line.
The output is the size of the largest blob found on the grid.
Sample Input
1
11000
01100
00101
10001
01011
Sample Output
5

#include <iostream>
#include <cstdlib>
#include <cstdio>
 
using namespace std;
 
char maps[30][30],buf[30];
 
int dxy[8][2] = {1,-1,1,0,1,1,0,-1,0,1,-1,-1,-1,0,-1,1};
int dfs(int x, int y)
{
	if (x < 0 || y < 0) return 0;
	if (maps[x][y] != '1') return 0;
	maps[x][y] = '0';
	int sum = 1;
	for (int k = 0 ; k < 8 ; ++ k)
		sum += dfs(x+dxy[k][0], y+dxy[k][1]);
	return sum;
}
 
int main()
{
	int n;
	scanf("%d",&n);
	gets(buf);
	gets(buf);
	while (n --) {
		int count = 0;
		while (gets(maps[count]))
			if(!maps[count ++][0]) break;
		
		int Max = 0;
		for (int i = 0 ; i < count ; ++ i)
		for (int j = 0 ; maps[i][j] ; ++ j)
			if (maps[i][j] == '1')
				Max = max(Max, dfs(i, j));
		
		printf("%d\n",Max);
		if (n) printf("\n");
	}
	
	return 0;
}