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.
InputWrite a program to count the number of black tiles which he can reach by repeating the moves described above.
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.
OutputThere 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.
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 Input6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0Sample Output
45 59 6 13
好久没做搜索的题目啦,直接用深搜做的,当然也可以用广搜来做,下面把代码贴在下面
DFS:
#include<iostream>
#include<cstdio>
#include<string>
#include<string.h>
using namespace std;
#define maxn 25
char mp[maxn][maxn];
int dir[4][2]={-1,0,1,0,0,-1,0,1},vis[maxn][maxn];
int w,h,cnt;
void dfs(int a,int b)
{
vis[a][b]=1;
cnt++;
for(int i=0;i<4;i++)
{
int x=a+dir[i][0];
int y=b+dir[i][1];
if(x>=0&&x<h&&y>=0&&y<w&&mp[x][y]=='.'&&!vis[x][y])
{
dfs(x,y);
}
}
}
int main()
{
int a,b;
while(cin>>w>>h){
if(w==0&&h==0)
break;
for(int i=0;i<h;i++)
{
for(int j=0;j<w;j++)
{
cin>>mp[i][j];
if(mp[i][j]=='@')
{
a=i;
b=j;
}
}
}
memset(vis,0,sizeof(vis));
cnt=0;
dfs(a,b);
cout<<cnt<<endl;
}
return 0;
}
BFS:
#include<iostream>
#include<cstdio>
#include<string>
#include<string.h>
#include<queue>
using namespace std;
#define maxn 25
char mp[maxn][maxn];
int dir[4][2]={-1,0,1,0,0,-1,0,1},vis[maxn][maxn];
int w,h,cnt;
typedef struct Node{
int m,n;
}Node;
void bfs(int a,int b)
{
queue<Node> q;
Node tmp,p;
vis[a][b]=1;
p.m=a,p.n=b;
q.push(p);
while(!q.empty()){
tmp=q.front();
cnt++;
q.pop();
for(int i=0;i<4;i++)
{
int x=tmp.m+dir[i][0];
int y=tmp.n+dir[i][1];
if(x>=0&&x<h&&y>=0&&y<w&&mp[x][y]=='.'&&!vis[x][y])
{
p.m=x,p.n=y;
q.push(p);
vis[x][y]=1;
}
}
}
}
int main()
{
int a,b;
while(cin>>w>>h){
if(w==0&&h==0)
break;
for(int i=0;i<h;i++)
{
for(int j=0;j<w;j++)
{
cin>>mp[i][j];
if(mp[i][j]=='@')
{
a=i;
b=j;
}
}
}
memset(vis,0,sizeof(vis));
cnt=0;
bfs(a,b);
cout<<cnt<<endl;
}
return 0;
}