zoj 2412 Farm Irrigation [ dfs ]

题目链接 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1412

大致题意：按照题目给出的几个正方形拼成一个矩形，每个小正方形中间有个管子，问拼成的矩形有多少个能通的管子。

思路：ｄｆｓ

#include<stdio.h>
#include<string.h>

#define I (i*3)
#define J (j*3)
char str[51][51];
int map[159][159];
int n, m;
int dir[4][2] = 
{
	{0, 1}, {0, -1}, {-1, 0}, {1, 0}
};

void dfs(int i, int j)
{
	map[i][j] = 0;
	for(int k = 0; k < 4; k++) {
		int x = i + dir[k][0];
		int y = j + dir[k][1];
		if(x >= 0 && x < 3*n && y >= 0 && y < 3*m && map[x][y]) {
			dfs(x, y);
		}
	}
}

int main()
{
	while(scanf("%d %d%*c", &n, &m) != EOF) {
		if(n < 0 || m < 0) break;
		memset(map, 0, sizeof(map));
		for(int i = 0; i < n; i++) {
			for(int j = 0; j < m; j++) {
				scanf("%c", &str[i][j]);
				if( str[i][j] == 'A' ) {
					map[I][J+1] = map[I+1][J] = map[I+1][J+1] = 1;
				}
				if( str[i][j] == 'B' ) {
					map[I+1][J+1] = map[I+1][J+2] = map[I][J+1] = 1;
				}
				if( str[i][j] == 'C' ) {
					map[I+2][J+1] = map[I+1][J] = map[I+1][J+1] = 1;
				}
				if( str[i][j] == 'D' ) {
					map[I+2][J+1] = map[I+1][J+2] = map[I+1][J+1] = 1;
				}
				if( str[i][j] == 'E' ) {
					map[I][J+1] = map[I+2][J+1] = map[I+1][J+1] = 1;
				}
				if( str[i][j] == 'F' ) {
					map[I+1][J+2] = map[I+1][J] = map[I+1][J+1] = 1;
				}
				if( str[i][j] == 'G' ) {
					map[I+1][J+2] = map[I+1][J] = map[I+1][J+1] = map[I][J+1] = 1;
				}
				if( str[i][j] == 'H' ) {
					map[I][J+1] = map[I+2][J+1] = map[I+1][J+1] = map[I+1][J] = 1;
				}
				if( str[i][j] == 'I' ) {
					map[I+1][J+2] = map[I+1][J] = map[I+1][J+1] = map[I+2][J+1] = 1;
				}
				if( str[i][j] == 'J' ) {
					map[I][J+1] = map[I+2][J+1] = map[I+1][J+1] = map[I+1][J+2] = 1;
				}
				if( str[i][j] == 'K' ) {
					map[I][J+1] = map[I+2][J+1] = map[I+1][J+1] = map[I+1][J+2] = map[I+1][J] = 1;
				}
			}
			scanf("%*c");
		}
		/**************************
		for(int i = 0; i < 3*n; i++) {
			for(int j = 0; j < 3*m; j++) {
				printf("%d ", map[i][j]);
			}
			printf("\n");
		}
		**************************/
		
		int ans = 0;
		for(int i = 0; i < 3*n; i++) {
			for(int j = 0; j < 3*m; j++) {
				if(map[i][j]) {
					dfs(i, j);
					ans++;
				}
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}