uva 705 - Slash Maze

开始无从下手,后来参考了别人的思路 http://blog.csdn.net/shuangde800/article/details/7726620

有了思路后不知道怎么求环,后来发现题目说了没岔路。。

 

#include<stdio.h>
#include<string.h>
#define MAX 200

int w, h;
char maze[MAX][MAX];
int cases = 1;
int isCircle;
int num;
int circleCount;
int maxLen;
int hasCircle;

int dir[8][2] = { { -1, 0 }, { 1, 0 }, { 0, -1 }, { 0, 1 }, { -1, -1 },
		{ -1, 1 }, { 1, -1 }, { 1, 1 } };

int pass(int i, int j, int k) {
	if (k < 4)
		return 1;
	int x = i + dir[k][0];
	int y = j + dir[k][1];
	if (x >= 0 && y >= 0 && maze[x][j] == '\\' && maze[i][y] == '\\'
			&& (k == 5 || k == 6))
		return 0;

	if (x >= 0 && y >= 0 && maze[x][j] == '/' && maze[i][y] == '/'
			&& (k == 4 || k == 7))
		return 0;

	return 1;

}

void dfs(int i, int j) {
	if (i < 0 || j < 0 || i >= 2 * h || j >= 2 * w) {
		isCircle = 0;
		return;
	}
	if (maze[i][j] == '*' || maze[i][j] == '\\' || maze[i][j] == '/')
		return;

	maze[i][j] = '*';
	num++;

	int k;
	for (k = 0; k < 8; k++) {
		if (pass(i, j, k)) {
			dfs(i + dir[k][0], j + dir[k][1]);
		}

	}

}

int main() {

	while (scanf("%d%d", &w, &h) != EOF) {
		if (w == 0)
			break;
		memset(maze, 0, sizeof(maze));
		getchar();
		int i, j;
		char ch;
		for (i = 0; i < h; i++) {
			for (j = 0; j < w; j++) {
				scanf("%c", &ch);
				if (ch == '\\') {
					maze[2 * i][2 * j] = '\\';
					maze[2 * i + 1][2 * j + 1] = '\\';
				} else if (ch == '/') {
					maze[2 * i][2 * j + 1] = '/';
					maze[2 * i + 1][2 * j] = '/';
				}

			}
			getchar();

		}

		circleCount = 0;
		maxLen = -1;
		hasCircle = 0;
		for (i = 0; i < 2 * h; i++)
			for (j = 0; j < 2 * w; j++) {
				if (maze[i][j] == 0) {
					num = 0;
					isCircle = 1;
					dfs(i, j);

					if (isCircle && num >= 4) {
						hasCircle = 1;
						circleCount++;
						if (num > maxLen)
							maxLen = num;
					}

				}
			}

		printf("Maze #%d:\n", cases++);
		if (hasCircle) {
			printf("%d Cycles; the longest has length %d.\n\n", circleCount,
					maxLen);
		} else {
			printf("There are no cycles.\n\n");
		}

	}

	return 0;
}

 

 

 

 

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