NEFU561 block calculation "recursion"

Topic Links:

http://acm.nefu.edu.cn/JudgeOnline/problemshow.php?problem_id=561

Main topic:

In a m*n rectangular room, a square tile with white and black color on the ground, you stand in one piece

On black tiles, you can only move to adjacent black tiles. Q: How many fast black tiles can be reached in total.

data, '. ' Represents a black tile, ' # ' represents a white tile, ' @ ' means the tile you stand on (the tile is black).

Ideas:

Can only move to adjacent black tiles, then for position (x, y), only to (X+1,y), (x,y+1), (X-1,y),

(x,y+1) The black tile moves. Every time you move, look at the tiles you haven't traversed, if it's a black tile,

go on, or you'll return. The number of black tile blocks traversed with ans, the final recursive equation is f (x, y) = f(x+1,y) +

(x,y+1) + (x-1,y) + (x,y+1).

To avoid repeating the number of tiles traversed, mark each piece of tile passed as white so that it does not repeat the calculation.

AC Code:

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring>using namespace Std;int N,m;char map[35][35];int DFS (int x,int y) {    int d = 0;    if (x < 0 | | x >= N | | y < 0 | | y >= M | | Map[x][y] = = ' # ')        return 0;   Return, recursive boundary    if (map[x][y] = = '. ' | | Map[x][y] = = ' @ ')    {        d = 1;      tags, avoid repeated searches        map[x][y] = ' # ';    }    Return Dfs (x+1,y) + DFS (x,y+1) + DFS (x-1,y) + DFS (x,y-1) + D;} int main () {    while (~scanf ("%d%d", &m,&n) && (n| | M))    {        int ans = 0;        for (int i = 0; i < N; ++i)            scanf ("%s", Map[i]);        for (int i = 0, i < N; ++i) for            (int j = 0; j < M; ++j)                if (map[i][j] = = ' @ ')                    ans = DFS (i,j);        printf ("%d\n", ans);    }    return 0;}

NEFU561 Block calculates "recursion"