Poj 1979 & zoj 2165 red and black

Red and black

Time limit:1000 ms		Memory limit:30000 K
Total submissions:22409		Accepted:12100

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 ctions, respectively. W and H are not more than 20.

There are h more lines in the data set, each of which between des 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 shocould output a line which contains the number of tiles he can reach from the initial tile (including itself ).

Sample Input

6 9....#......#..............................#@...#.#..#.11 9.#..........#.#######..#.#.....#..#.#.###.#..#.#[email protected]#.#..#.#####.#..#.......#..#########............11 6..#..#..#....#..#..#....#..#..###..#..#..#@...#..#..#....#..#..#..7 7..#.#....#.#..###.###[email protected]###.###..#.#....#.#..0 0

Sample output

4559613

 

 

 

A simple DFS question that counts the number of searchable items. A man is standing on a black tile. There are black and red tiles next to him, but he can only walk in the black tile, and there are four directions, ask him how many black tiles he can use at most.

 

#include <stdio.h>#include <string.h>#include <iostream>using namespace std;int num;char map[25][25];int visited[25][25];int dir[4][2] = {0, -1, 1, 0, 0, 1, -1, 0};void DFS(int x, int y, int n, int m){    int mx, my;    for(int i = 0; i<4; i++)    {        mx = x+dir[i][0];    my = y+dir[i][1];        if(mx>=1 && mx<=n && my>=1 && my<=m && !visited[mx][my] && map[mx][my]==‘.‘)        {            visited[mx][my] = 1;            num++;            DFS(mx, my, n, m);        }    }}int main(){    int n, m, x, y;    while(scanf("%d%d", &m, &n)!=EOF && (n || m))    {        num = 1;        memset(visited, 0, sizeof(visited));        for(int i = 1; i<=n; i++)        {            for(int j = 1; j<=m; j++)            {                cin>>map[i][j];                if(map[i][j] == ‘@‘)                {                    x = i;                    y = j;                    visited[x][y] = 1;                }            }        }        DFS(x, y, n, m);        printf("%d\n", num);    }    return 0;}