Red and Black-----BFS(水题)
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!!!