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POJ 1979 Red and Black （DFS）

2024-11-26 03:11:39   203人阅读

Red and Black
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 23904 Accepted: 12927

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 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

Source

Japan 2004 Domestic



题意：给一个图，‘@‘是起点，‘.‘可以通过，‘#‘不可以通过。问从起点出发最多可以到达图中的多少个点。


解析：从起点深搜即可，每次dfs的时候，都把当前的起点改为’@‘，每次向四个方向搜索的时候，只有不出边界且当前点为’.‘的时候才继续深搜。只需要记录多少次dfs即为答案。




AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <stack>
using namespace std;
#define INF 0x7fffffff
#define LL long long
#define MID(a, b)  a+(b-a)/2
const int maxn = 1000 + 10;
int d[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};
int m, n, ans;
char s[22][22];

void dfs(int x, int y){
    s[x][y] = '@';
    ans ++;
    for(int i=0; i<4; i++){
        int dx = x + d[i][0], dy = y + d[i][1];
        if(dx < n && dx >=0 && dy < m && dy >=0 && s[dx][dy] == '.')              //开始用的条件是s[dx][dy] != '#'，程序老是死掉！！！
            dfs(dx, dy);
    }
}

int main(){
    #ifdef sxk
        freopen("in.txt", "r", stdin);
    #endif // sxk

    int sx = 0, sy = 0;
    while(scanf("%d%d", &m, &n)!=EOF  && !(!n && !m)){
        ans = 0;
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++){
                cin >> s[i][j];
                if(s[i][j] == '@'){                 //记录起点坐标
                    sx = i;
                    sy = j;
                }
            }
        dfs(sx, sy);
        printf("%d\n", ans);
    }
    return 0;
}




POJ 1979 Red and Black （DFS）


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