POJ-1979 Red and Black-----BFS(水题)

最新推荐文章于 2020-08-11 21:22:28 发布

原创最新推荐文章于 2020-08-11 21:22:28 发布 · 179 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#Red and Black #水题

BFS 专栏收录该内容

3 篇文章

订阅专栏

本文介绍了一个使用广度优先搜索(BFS)算法来计算在一个由黑色和红色方块组成的矩形房间中，从指定的初始位置可达的黑色方块数量的程序实现。通过给出的具体代码示例，展示了如何定义地图结构、执行BFS搜索并统计可达方块数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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
45
59
6
13

裸的BFS,直接套模板就行了，我也是小白，多多包涵，代码如下：

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
const int N=30;
int n,m;
int ex,ey;
int bj[N][N];
char tu[N][N];
struct peo{
	int x,y;
};
queue<peo> que;
int d[4][2]={1,0,-1,0,0,1,0,-1};
void bfs(int x,int y)
{
	memset(bj,0,sizeof(bj));
	peo t,tt;
	t.x=x;
	t.y=y;
	que.push(t);
	bj[x][y]=1;
	while(!que.empty())
	{
		t=que.front();
		que.pop();
		for(int i=0;i<4;i++){
			int dx=t.x+d[i][0];tt.x=dx;
			int dy=t.y+d[i][1];tt.y=dy;
			if(tu[dx][dy]=='#'||dx<0||dx>=m||dy<0||dy>=n||bj[dx][dy])continue;
			que.push(tt);
			bj[dx][dy]=1;
		}
	}
}
int main()
{
	
	while(~scanf("%d%d",&n,&m)&&n&&m)
	{
		for(int i=0;i<m;i++)
		for(int j=0;j<n;j++)
		{
			cin>>tu[i][j];
			if(tu[i][j]=='@')ex=i,ey=j;
		}
		bfs(ex,ey);
		int ans=0;
		for(int i=0;i<m;i++)
		for(int j=0;j<n;j++)
		ans+=bj[i][j];
		cout<<ans<<endl;
	}
	return 0;
}

欢迎任何形式的转载，但是请标注来源！！！