HDU-1312-Red and Black（搜索）

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本文通过一个具体的房间路径搜索问题，介绍了如何使用宽度优先搜索(BFS)算法来寻找可达路径的最大数量。文章详细展示了从起始位置出发，在遵循特定规则的情况下，如何遍历地图上的所有可达节点，并给出了完整的C++代码实现。

题目链接---HDU-Red and Black

Red and Black

Time Limit:1000MS Memory Limit:30000KB 64bit IO Format:%I64d & %I64u

Submit Status

Description

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)
The end of the input is indicated by a line consisting of two zeros.

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

题意：房间里（图中的矩阵）有红砖和黑砖，黑砖即' . '能踩，红砖' # '不能踩，初始位置在'@'位置，只能上下左右行走，遍历整张图求所踩的最大的砖数

题解：这道题是一道典型的搜索题，下面我用的是bfs解决的这道题

代码实现如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define MAX 25
#define INF 0x3f3f3f3f 
#include<queue>
using namespace std;
typedef pair<int,int>P;
char maps[MAX][MAX];
int dis[MAX][MAX]; 
int dx[]={1,0,0,-1};//表示方向 
int dy[]={0,1,-1,0};
char sx,sy;//表示开始坐标
int m,n;
queue<P>que;
int bfs()
{
	for(int i=0;i<m;i++)
	{
		for(int j=0;j<n;j++)
		{
			dis[i][j]=INF;
		}
	}
	que.push(P(sx,sy));
	dis[sx][sy]=0;
	int k=1;//记录所走的砖数 
	while(que.size())
	{
		P s=que.front();
		que.pop();
		for(int i=0;i<4;i++)
		{
			int nx=s.first+dx[i];
			int ny=s.second+dy[i];
			if(nx>=0&&nx<m&&ny>=0&&ny<n&&dis[nx][ny]==INF&&maps[nx][ny]=='.')
			{
				que.push(P(nx,ny));
				dis[nx][ny]=dis[s.first][s.second]+1; 
				k++;
			}
		}
	}
	return k;
} 
int main()
{
	while(cin>>n>>m)
	{
		if(n==0&&m==0) break;
		for(int i=0;i<m;i++)
		{
		    cin>>maps[i]; 
		}
		for(int i=0;i<m;i++)
		{
			for(int j=0;j<n;j++)
			{
				if(maps[i][j]=='@')
				{
					sx=i;
					sy=j;
				}
			}
		}
		int ans=bfs();
		cout<<ans<<endl;
	}
return 0;	
}