POJ 1979 Red and Black

    xiaoxiao2025-02-10  18

    题目链接:http://poj.org/problem?id=1979

    Red and Black

    Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 31008 Accepted: 16914

    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 0

    Sample Output

    45 59 6 13

    Source

    Japan 2004 Domestic

    思路:最基本的DFS吧!直接上代码吧!

    附上AC代码:

    #include <cstdio> using namespace std; const int maxn = 25; const int dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}}; char mp[maxn][maxn]; int n, m, ans; void dfs(int x, int y){ if (x<0 || x>=n || y<0 || y>=m) return ; if (mp[x][y] == '#') return ; mp[x][y] = '#'; ++ans; for (int i=0; i<4; ++i) dfs(x+dir[i][0], y+dir[i][1]); } int main(){ while (~scanf("%d%d", &m, &n) && n+m){ for (int i=0; i<n; ++i) scanf("%s", mp[i]); ans = 0; int x=0, y=0; for (int i=0; i<n; ++i) for (int j=0; j<m; ++j) if (mp[i][j] == '@'){ x = i; y = j; break; } dfs(x, y); printf("%d\n", ans); } return 0; }

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