<span style="color:#330099;">/*
B - 广搜/深搜 基础
Time Limit:1000MS Memory Limit:30000KB 64bit IO Format:%I64d & %I64u
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Status
Description
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
By Grant Yuan
2014.7.12
*/
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<queue>
#include<cstdio>
using namespace std;
bool flag[21][21];
char a[21][21];
int n,m;
int s1,f1;
typedef struct{
int x;
int y;
}node;
int next[4][2]={1,0,0,1,-1,0,0,-1};
int sum;
queue<node>q;
bool can(int cc,int dd)
{
if(cc>=0&&cc<m&&dd>=0&&dd<n&&flag[cc][dd]==0)
return 1;
return 0;
}
void slove()
{ int c,d,cc,dd;
node q1;
while(!q.empty()){
c=q.front().x;d=q.front().y;
for(int i=0;i<4;i++)
{
cc=c+next[i][0];
dd=d+next[i][1];
if(can(cc,dd))
{ q1.x=cc;
q1.y=dd;
q.push(q1);
flag[cc][dd]=1;
sum++;
}
}
q.pop();
}
}
int main()
{ node q1;
while(1){
sum=1;
memset(flag,0,sizeof(flag));
cin>>n>>m;
if(n==0&&m==0)
break;
for(int i=0;i<m;i++){
scanf("%s",&a[i]);}
for(int i=0;i<m;i++)
for(int j=0;j<n;j++)
{if(a[i][j]=='#')
flag[i][j]=1;
if(a[i][j]=='@')
s1=i,f1=j;
}
q1.x=s1;
q1.y=f1;
q.push(q1);
flag[s1][f1]=1;
slove();
cout<<sum<<endl;
}
return 0;
}
</span>原文：http://blog.csdn.net/yuanchang_best/article/details/37730447