poj 1979 Red and Black 题解《挑战程序设计竞赛》

地址 http://poj.org/problem?id=1979

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 .

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

....#.
.....#
......
......
......
......
......
#@...#
.#..#.

.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........

..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..

..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..

Sample Output

大概意思就是 @是一个人的起点 可以在.的地板砖上移动 但是不能移动到#的地板上。求能够达到的最多地板
算是基础的dfs题目 没有剪枝 就是遍历

代码

#include <iostream>

using namespace std;

int n,m;

const int N = ;

char graph[N][N];

char visit[N][N];
int ret = ;

int addx[] = { ,-,, };
int addy[] = { ,,,- };

void Dfs(int x, int y) {
    if (x <  || x >= n || y <  || y >= m) return;

    if (visit[x][y] ==   || graph[x][y] == '#') return;

    visit[x][y] = ;
    if (graph[x][y] == '.') {
        //cout << "x = " << x << ".   y = " << y << endl;
        ret++;
    }

    for (int i = ; i < ; i++) {
        int newx = x + addx[i];
        int newy = y + addy[i];
        Dfs(newx, newy);
    }

    return;
}

int main()
{
    while (cin >> m >> n) {
        if (m ==  || n == ) break;
        int x, y;
        ret = ;
        memset(visit, , N*N);

        for (int i = ; i < n; i++) {
            for (int j = ; j < m; j++) {
                cin >> graph[i][j];
                if (graph[i][j] == '@') {
                    x = i; y = j;
                    ret++;
                }
            }
        }

        Dfs(x, y);

        cout << ret << endl;
    }

    //system("pause");

    return ;
}