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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 | 4559613 |
【题解】
数据不大，DFS可解。
【代码 C++】
1 #include <cstdio> 2 #include <cstring> 3 char data[25][25]; 4 int sum; 5 void DFS(int y, int x){ 6 if (data[y][x] == ‘#‘) return; 7 ++sum; data[y][x] = ‘#‘; 8 DFS(y + 1, x); DFS(y - 1, x); 9 DFS(y, x + 1); DFS(y, x - 1);10 }11 int main(){12 int w, h, i, j, stY, stX;13 while (~scanf("%d%d ", &w, &h)){14 if (w + h == 0) break;15 memset(data, ‘#‘, sizeof(data));16 for (i = 1; i <= h; ++i){17 gets(&data[i][1]);18 for (j = 1; j <= w; ++j) if (data[i][j] == ‘@‘) stY = i, stX = j;19 data[i][j] = ‘#‘;20 }21 sum = 0; DFS(stY, stX);22 printf("%d\n", sum);23 }24 return 0;25 }
POJ 1979 Red and Black （红与黑）