POJ 1979 Red and Black （红与黑）

POJ 1979 Red and Black （红与黑）

Time Limit: 1000MS    Memory Limit: 30000K

 

Description	题目描述
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.	有个铺满方形瓷砖的矩形房间，每块瓷砖的颜色非红即黑。某人在一块砖上，他可以移动到相邻的四块砖上。但他只能走黑砖，不能走红砖。   敲个程序统计一下这样可以走到几块红砖上。

 

Input

输入

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.

多组测试用例。每组数组开头有两个正整数W和H；W与H分别表示 x- 与 y- 方向上瓷砖的数量。W和W均不超过20。   还有H行数据，每行包含W个字符。每个字符表示各色瓷砖如下。   ‘.’- 一块黑砖 ‘#’- 一块红砖 ‘@’- 一个黑砖上的人（一组数据一个人） 输入以一行两个零为结束。

 

Output	输出
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).	对于每组测试用例，输出他从起始砖出发所能抵达的瓷砖数量（包括起始砖）。

 

Sample Input - 输入样例	Sample Output - 输出样例
6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0	45 59 6 13

 

【题解】

　　数据不大，DFS可解。

【代码 C++】

#include <cstdio>
#include <cstring>
char data[25][25];
int sum;
void DFS(int y, int x){
    if (data[y][x] == '#') return;
    ++sum; data[y][x] = '#';
    DFS(y + 1, x); DFS(y - 1, x);
    DFS(y, x + 1); DFS(y, x - 1);
}
int main(){
    int w, h, i, j, stY, stX;
    while (~scanf("%d%d ", &w, &h)){
        if (w + h == 0) break;
        memset(data, '#', sizeof(data));
        for (i = 1; i <= h; ++i){
            gets(&data[i][1]);
            for (j = 1; j <= w; ++j) if (data[i][j] == '@') stY = i, stX = j;
            data[i][j] = '#';
        }
        sum = 0; DFS(stY, stX);
        printf("%d\n", sum);
    }
    return 0;
}