poj1741，POJ1164 The Castle【DFS】

The Castle
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 8317 Accepted: 4693

Description

poj1741，1 2 3 4 5 6 7

#############################

1 # | # | # | | #

#####—#####—#---#####—#

2 # # | # # # # #

#—#####—#####—#####—#

3 # | | # # # # #

#—#########—#####—#---#

4 # # | | | | # #

#############################

(Figure 1)

# = Wall

| = No wall

- = No wall

Figure 1 shows the map of a castle.Write a program that calculates

1. how many rooms the castle has
2. how big the largest room is
   The castle is divided into m * n (m<=50, n<=50) square modules. Each such module can have between zero and four walls.

Input

Your program is to read from standard input. The first line contains the number of modules in the north-south direction and the number of modules in the east-west direction. In the following lines each module is described by a number (0 <= p <= 15). This number is the sum of: 1 (= wall to the west), 2 (= wall to the north), 4 (= wall to the east), 8 (= wall to the south). Inner walls are defined twice; a wall to the south in module 1,1 is also indicated as a wall to the north in module 2,1. The castle always has at least two rooms.

Output

Your program is to write to standard output: First the number of rooms, then the area of the largest room (counted in modules).

Sample Input

4
7
11 6 11 6 3 10 6
7 9 6 13 5 15 5
1 10 12 7 13 7 5
13 11 10 8 10 12 13

Sample Output

5
9

Source

IOI 1994

問題鏈接：POJ1164 The Castle
問題簡述：（略）
問題分析：
????這是一個搜索問題，用DFS實現。
程序說明：（略）
參考鏈接：（略）
題記：（略）

AC的C語言程序如下：

/* POJ1164 The Castle */#include <stdio.h>int drow[] = {0, -1, 0, 1};
int dcol[] = {-1, 0, 1, 0};#define N 50
int maze[N][N], n, m, sum, cnt;void dfs(int row, int col)
{int s = maze[row][col], i;sum++;maze[row][col] = -1;for(i = 0; i < 4; i++) {if(s % 2 == 0) {int nrow = row + drow[i];int ncol = col + dcol[i];if(nrow >= 0 && nrow < n && ncol >= 0 && ncol < m && maze[nrow][ncol] != -1)dfs(nrow, ncol);}s /= 2;}
}int main(void)
{int maxsum, i, j;while(scanf("%d%d", &n, &m) != EOF) {for(i = 0; i < n; i++)for(j = 0; j < m; j++)scanf("%d", &maze[i][j]);cnt = maxsum = 0;for(i = 0; i < n; i++)for(j = 0; j < m; j++)if(maze[i][j] != -1) {sum = 0;dfs(i, j);cnt++;if(sum > maxsum)maxsum = sum;}printf("%d\n%d\n", cnt, maxsum);}return 0;
}