Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15951 Accepted Submission(s): 9849
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.
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)
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
int w,h,cnt;
char mp[21][21];
struct pos{
int x,y;
}st;
bool vis[21][21];
bool ok(int x,int y){
if(x<0 || x>=h)return false;
if(y<0 || y>=w)return false;
return true;
}
void bfs(){
int mve[][2]={1,0, -1,0, 0,1, 0,-1};
queue<pos> qu;
qu.push(st);
pos cur,next;
memset(vis, false, sizeof(vis));
vis[st.x][st.y] = true;
while(!qu.empty()){
cur = qu.front();qu.pop();
for(int i=0;i<4;i++){
int nx = cur.x+mve[i][0];
int ny = cur.y+mve[i][1];
if(!ok(nx,ny) || vis[nx][ny] || mp[nx][ny]=='#')continue;
vis[nx][ny]=true;
cnt++;
next.x=nx,next.y=ny;
qu.push(next);
}
}
}
int main(){
//freopen("hdu1312.in", "r", stdin);
//freopen("hdu.out", "w", stdout);
while(cin>>w>>h && w){
int flag=1;
cnt=1;
for(int i=0;i<h;i++){
scanf("%s", mp[i]);
if(1==flag)
for(int j=0;j<w;j++)
if(mp[i][j]=='@')
st.x=i,st.y=j;
}
bfs();
cout<<cnt<<endl;
}
return 0;
}