题目
Red and Black
Time Limit: 1000MS Memory Limit: 30000K
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
题目大意
有一个人在走一个迷宫，他每次只能走上下左右中的一个方位，有一些障碍不能走，求其最多能走的位置个数。
分析
模板题，用bfs即可
代码
#include <iostream>
#include <queue>
using namespace std;
char op[100];
bool st[100][100];
int dx[4] = {0,0,1,-1},dy[4] = {-1,1,0,0};
int main()
{int m,n,x,y;while(scanf("%d%d",&m,&n) && m && n){int cnt = 0;for(int i = 0; i < n; i ++){scanf("%s",op);for(int j = 0; j < m; j ++)if(op[j] == '.')st[i][j] = true;else if(op[j] == '@'){x = i, y = j;st[i][j] = false; }else st[i][j] = false;}pair<int,int> now[1000];now[1] = make_pair(x,y);int idx = 1;queue<int> q;q.push(1);cnt ++;while(!q.empty()){int t = q.front();q.pop();for(int i = 0; i < 4; i ++){int tx = now[t].first + dx[i], ty = now[t].second + dy[i];if(tx >= 0 && tx < n && ty>=0 && ty < m && st[tx][ty]){st[tx][ty] = false;now[++idx] = make_pair(tx,ty);q.push(idx);cnt ++;}}}printf("%d\n",cnt);}return 0;
}