原题传送门：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2165
Red and Black
Time Limit: 2 Seconds Memory Limit: 65536 KB
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
Source: Asia 2004, Ehime (Japan), Japan Domestic
这道题主要是考察BFS（广度优先搜索），每次去搜索与某个点相邻的（上下左右）四个点，如果这四个点中有符合条件---1.x>0,y>0,x<row,y<cloumn,并且没有被访问过，以及不是红色砖块。那么将这个个点入队。那么用什么条件来控制这个循环呢？答案当然是队列为空！
首先将起始点Pos，进队，然后while(队列不为空){ count++; 检查四周有木有符合上述条件的点，有则入队} printf count is ok!
//图的BFS ZOJ 2165
#include <stdio.h>
#include <queue>
#include <string.h>
using namespace std;
struct Pos{
int x;
int y;
};
const int MAXN = 21;
char list[MAXN][MAXN];
bool isVisited[MAXN][MAXN];
int row,cl,count;
Pos staPos;
int main(){
queue<Pos> bfs;
while(scanf("%d %d",&cl,&row)!= EOF){
if(cl == 0&&row==0)break;
count = 0;
for(int i=1;i<MAXN;i++){
for(int j=1;j<MAXN;j++){
isVisited[i][j] = false;
}
}
for(int i=1;i<=row;i++){
scanf("%s",list[i]+1);
}
for(int i=1;i<=row;i++){
for(int j=1;j<=cl;j++){
if(list[i][j] == '@'){
staPos.x = i;
staPos.y = j;
}
}
}
bfs.push(staPos);
isVisited[staPos.x][staPos.y] = true;
while(!bfs.empty()){
count++;
Pos tempPos = bfs.front();
bfs.pop();
if((list[tempPos.x+1][tempPos.y] == '.')&& (tempPos.x+1<=row) && (!isVisited[tempPos.x+1][tempPos.y])){
Pos newPos1;
newPos1.x = tempPos.x+1;
newPos1.y = tempPos.y;
bfs.push(newPos1);
isVisited[newPos1.x][newPos1.y] = true;
}
if((list[tempPos.x-1][tempPos.y] == '.')&& (tempPos.x-1>0) && (!isVisited[tempPos.x-1][tempPos.y])){
Pos newPos2;
newPos2.x = tempPos.x-1;
newPos2.y = tempPos.y;
bfs.push(newPos2);
isVisited[newPos2.x][newPos2.y] = true;
}
if((list[tempPos.x][tempPos.y+1] == '.')&& (tempPos.y+1<=cl) && (!isVisited[tempPos.x][tempPos.y+1])){
Pos newPos3;
newPos3.x = tempPos.x;
newPos3.y = tempPos.y+1;
bfs.push(newPos3);
isVisited[newPos3.x][newPos3.y] = true;
}
if((list[tempPos.x][tempPos.y-1] == '.')&& (tempPos.y-1>0) && (!isVisited[tempPos.x][tempPos.y-1])){
Pos newPos4;
newPos4.x = tempPos.x;
newPos4.y = tempPos.y-1;
bfs.push(newPos4);
isVisited[newPos4.x][newPos4.y] = true;
}
}
printf("%d\n",count);
}
return 0;
}