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
45 59 6 13
C++版本一
BFS
#include <iostream>
#include <stdio.h>
#include <cstring>
#include <queue>
using namespace std;
int t,m,n,ans;
int k,x1,x2,y1,y2;
char dfs[110][110];
//int vis[110][110];
int a[4]={1,-1,0,0};
int b[4]={0,0,1,-1};
struct node
{
int x;
int y;
}front1,tmp,start;
queue <node>Q;
bool prime(int x,int y){
if(tmp.x>=0&&tmp.x<n&&tmp.y>=0&&tmp.y<m)
return true;
else
return false;
}
void bfs(node s){
while(!Q.empty())
Q.pop();
Q.push(s);
while(!Q.empty()){
front1=Q.front();
Q.pop();
for(int i=0;i<4;i++){
tmp.x=a[i]+front1.x;
tmp.y=b[i]+front1.y;
if(prime(tmp.x,tmp.y)&&dfs[tmp.x][tmp.y]!='#'){
dfs[tmp.x][tmp.y]='#';
// printf("%d %d %d\n",tmp.x,tmp.y,i);
ans++;
Q.push(tmp);
}
}
}
}
int main()
{
while (scanf("%d%d",&m,&n)!=EOF) {
if(m==0&&n==0)break;
// memset(vis,0,sizeof(vis));
memset(dfs,'0',sizeof(dfs));
getchar();
for(int i=0;i<n;i++){
scanf("%s",dfs[i]);
for(int j=0;j<m;j++){
if(dfs[i][j]=='@')
{
x1=i;
y1=j;
break;
}
}
getchar();
}
start.x=x1;
start.y=y1;
dfs[x1][y1]='#';
ans=1;
bfs(start);
printf("%d\n",ans);
}
return 0;
}
C++版本二
DFS
#include<iostream>
#include <cstdio>//getchar()头文件
using namespace std;
char a[25][25];
int dir[4][2]=
{
{1,0}, //向右
{-1,0},//向左
{0,1}, //向上
{0,-1} //向下
};
int x,y,num;
bool YES(int x0,int y0)//xx,yy
{
//x--列,y--行
if(x0<y&&x0>=0&&y0>=0&&y0<x)
return true;
else
return false;
}
void DFS(int x0,int y0)
{
a[x0][y0]='#';
num++;
int xx,yy;
for(int i=0; i<4; i++)
{
int xx=x0+dir[i][0];
int yy=y0+dir[i][1];
if(YES(xx,yy)&&a[xx][yy]=='.')
DFS(xx,yy);
}
}
int main()
{
int i,j,di,dj;
while (scanf("%d%d",&x,&y)!=EOF)
{
getchar();
if (x==0&&y==0) break;
for (i=0; i<y; i++)
{
for (j=0; j<x; j++)
{
scanf("%c",&a[i][j]);
if(a[i][j] == '@')
{
di = i;
dj = j;
}
}
getchar();
}
num=0;
DFS(di,dj);
cout<<num<<endl;
}
return 0;
}
C++版本三
BFS
#include<cstdio>
#include<queue>
using namespace std;
char a[25][25];
int x,y,sum;
int dir[4][2]={1,0,0,1,-1,0,0,-1};
struct node
{
int x,y;
}q[500];
bool YES(int x0,int y0)//xx,yy
{
//x--列,y--行
if(x0<y&&x0>=0&&y0>=0&&y0<x)
return true;
else
return false;
}
void BFS(int x0,int y0)
{
queue<node>q;
node start,endd;
start.x=x0;
start.y=y0;
q.push(start);
while(!q.empty())
{
start=q.front();
q.pop();
for(int i=0;i<4;i++)
{
endd.x=start.x+dir[i][0];
endd.y=start.y+dir[i][1];
if(YES(endd.x,endd.y)&&a[endd.x][endd.y]=='.')
{
a[endd.x][endd.y]='#';
sum++;
q.push(endd);
}
}
}
}
int main()
{
int i,j,di,dj;
while (scanf("%d%d",&x,&y)!=EOF)
{
getchar();
if (x==0&&y==0) break;
for (i=0; i<y; i++)
{
for (j=0; j<x; j++)
{
scanf("%c",&a[i][j]);
if(a[i][j] == '@')
{
di = i;
dj = j;
}
}
getchar();
}
sum=1;
BFS(di,dj);
printf("%d\n",sum);
}
return 0;
}
C语言版本一
BFS
#include<cstdio>
#include<queue>
using namespace std;
int a[25][25];
int x,y,sum;
int dir[4][2]={1,0,0,1,-1,0,0,-1};
struct node
{
int x,y;
}q[500];
bool YES(int x0,int y0)//xx,yy
{
//x--列,y--行
if(x0<y&&x0>=0&&y0>=0&&y0<x)
return true;
else
return false;
}
void BFS(int x0,int y0)
{
int head=0,tail=1;
q[1].x=x0;
q[1].y=y0;
while(head<tail)
{
++head;
for(int i=0;i<4;i++)
{
int x1=q[head].x+dir[i][0];
int y1=q[head].y+dir[i][1];
if(YES(x1,y1)&&a[x1][y1]=='.')
{
a[x1][y1]='#';
sum++;
q[++tail].x=x1;
q[tail].y=y1;
}
}
}
}
int main()
{
int i,j,di,dj;
while (scanf("%d%d",&x,&y)!=EOF)
{
getchar();
if (x==0&&y==0) break;
for (i=0; i<y; i++)
{
for (j=0; j<x; j++)
{
scanf("%c",&a[i][j]);
if(a[i][j] == '@')
{
di = i;
dj = j;
}
}
getchar();
}
sum=1;
BFS(di,dj);
printf("%d\n",sum);
}
return 0;
}
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