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 0Sample Output
45 59 6 13
题意:一个地图,有障碍,问从指定起始位置能到多少不同位置。
题解:随意地图题,dfs或者bfs。
AC代码:
#include<string.h> #include<stdio.h> #include<queue> using namespace std; int map[30][30]; int main() { char c; queue<int>x, y; int w, h,ans; int x0, y0; while (scanf("%d%d", &w, &h) != EOF&&w) { while (x.empty() != 1) { x.pop();y.pop(); } memset(map, 0, sizeof(map)); for (int i = 1;i <= h; i++) { getchar(); for (int j = 1;j <= w;j++) { c = getchar(); if (c == '#')map[i][j] = 1; if (c == '@') { x.push(i);y.push(j); } } } while (x.size()) { x0 = x.front(); y0 = y.front(); x.pop(); y.pop(); map[x0][y0] = -1; if (x0 - 1 >= 1 && map[x0 - 1][y0] == 0) { x.push(x0 - 1);y.push(y0); } if (y0 - 1 >= 1 && map[x0][y0 - 1] == 0) { x.push(x0);y.push(y0 - 1); } if (x0 + 1 <= h && map[x0 + 1][y0] == 0) { x.push(x0 + 1);y.push(y0); } if (y0 + 1 <= w && map[x0][y0 + 1] == 0) { x.push(x0);y.push(y0 + 1); } } ans = 0; for (int i = 1;i <= h;i++) for (int j = 1;j <= w;j++) if (map[i][j] == -1)ans++; printf("%d\n", ans); } }