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 |
45 |
【题解】
数据不大，DFS可解。
【代码 C++】
#include <cstdio>
#include <cstring>
char data[][];
int sum;
void DFS(int y, int x){
if (data[y][x] == '#') return;
++sum; data[y][x] = '#';
DFS(y + , x); DFS(y - , x);
DFS(y, x + ); DFS(y, x - );
}
int main(){
int w, h, i, j, stY, stX;
while (~scanf("%d%d ", &w, &h)){
if (w + h == ) break;
memset(data, '#', sizeof(data));
for (i = ; i <= h; ++i){
gets(&data[i][]);
for (j = ; j <= w; ++j) if (data[i][j] == '@') stY = i, stX = j;
data[i][j] = '#';
}
sum = ; DFS(stY, stX);
printf("%d\n", sum);
}
return ;
}