HDUOJ1312 Red and Black

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2083    Accepted Submission(s): 1356

Problem DescriptionThere 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. 

 

InputThe 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)  

OutputFor 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

4559613

 

【解題思路】像四個方向搜尋可達的.的數量，並吧訪問過的標記為#

 

 

 

 

#include<iostream><br />#include<stdio.h><br />using namespace std;<br />#define N 21<br />char arc[N][N];<br />int set[N][N], w, h;<br />int d[4][2] = { { -1, 0 }, { 1, 0 }, { 0, 1 }, { 0, -1 } };<br />int k;<br />int dfs(int x, int y) {<br />arc[x][y] = '#';<br />for (int i = 0; i < 4; i++) {<br />int px = x + d[i][0];<br />int py = y + d[i][1];<br />if (px >= 0 && py >= 0 && px < h && py < w && arc[px][py] != '#') {<br />k++;<br />dfs(px, py);<br />}<br />}<br />return k;<br />}<br />int main() {<br />int i, j;<br />int x, y;<br />while (cin >> w >> h && w && h) {<br />k = 1;<br />for (i = 0; i < h; i++)<br />for (j = 0; j < w; j++) {<br />cin >> arc[i][j];<br />if (arc[i][j] == '@') {<br />x = i;y = j;<br />}<br />}<br />cout << dfs(x, y) << endl;<br />}<br />return 0;<br />}