HDU 1312 Red and Black DFS入门经典例题

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) 

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

4559613

这两天做的第四道DFS题了，，这题很水，但因为我在做题时的粗心改了段时间才改出来

错误一：将x方向的变量和y方向的变量混了，导致坐标都存错了

错误二：将全局变量作为参数传入了DFS函数，导致每回一次初始状态的DFS函数的调用累加起来的变量瞬间归零。

错误三：储存点移动方向的数组名起错了，与某个库中函数名重复了。

贴代码：

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#define maxn 25using namespace std;int m, n, idex[maxn][maxn], nums;char field[maxn][maxn];int nextj[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};void DFS(int x, int y){    if(x < 0|| y < 0|| x >= n|| y >= m) return;    if(field[x][y] == '#' || idex[x][y] != 0) return;    nums++;    idex[x][y] = 1;    int xx, yy;    for(int i = 0; i < 4; i++)    {        xx = x + nextj[i][0], yy = y + nextj[i][1];        DFS(xx, yy);    }}int main(){    while(scanf("%d %d", &m, &n)&& m)    {        memset(idex, 0, sizeof(idex));        int x, y;        nums = 0;        for(int i = 0; i < n; i++)        {            for(int j = 0; j < m; j++)            {                cin >> field[i][j];                if(field[i][j] == '@')                {                    x = i, y = j;                }            }        }        DFS(x, y);        cout << nums << endl;    }    return 0;}