Red and Black//POJ - 1979//dfs

Red and Black//POJ - 1979//dfs

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题目

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
题意
与@连起来的点的数量
链接:http://poj.org/problem?id=1979

思路

dfs模板题。

代码

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <string>

using namespace std;
char s[50][50];
int cnt=0;
int l[4]={-1,0,1,0};
int r[4]={0,1,0,-1};
void dfs(int i,int j){
    if(s[i][j]=='#') return;
    else if(s[i][j]=='.'||s[i][j]=='@') {
        cnt++;
        s[i][j]='#';
        for(int t=0;t<4;t++){
            if(i+l[t]>=0&&j+r[t]>=0)
                dfs(i+l[t],j+r[t]);
        }
    }
}
int main()
{
    int n,m;
    while(1){
        cin>>m>>n;
        if(m==0&&n==0) break;
        memset(s,0,sizeof(s));
        for(int i=0;i<n;i++){
            scanf("%s",s[i]);
        }
        cnt=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(s[i][j]=='@'){
                    dfs(i,j);
                    goto A;
                }
            }
        }
        A:cout<<cnt<<endl;
    }
    return 0;
}

注意