题意：由'\' 和'/'组成的图，问是否有环的存在，如果有输出最大的环所占的格子数量。
题解：刚开始看没思路，在网上见到有人把斜线放大三倍，然后用1替代斜线，其他空白地方用0替代，然后普通的dfs再去写就很容易了。
#include <stdio.h>
#include <string.h>
const int N = 250;
int row, col, pos[N][N], flag, count, max, step;
int flag1[4] = {0, 0, 1, -1};
int flag2[4] = {1, -1, 0, 0};
void init() {
memset(pos, 0, sizeof(pos));
count = flag = step = 0;
max = -1;
}
void dfs(int x, int y) {
pos[x][y] = 1;
int x0, y0;
for (int i = 0; i < 4; i++) {
x0 = x + flag1[i];
y0 = y + flag2[i];
if (x0 < 0 || x0 >= row || y0 < 0 || y0 >= col) {
flag = 0;
continue;
}
else
if (pos[x0][y0] == 0) {
step++;
dfs(x0, y0);
}
}
}
int main() {
char c;
int t = 1;
while (scanf("%d%d", &col, &row) && (col || row)) {
getchar();
init();
int m = 0, n = 0;
for (int i = 0; i < row; i++, m += 3) {
for (int j = 0, n = 0; j < col + 1; j++, n += 3) {
scanf("%c", &c);
if (c == '\\') {
pos[m][n] = 1;
pos[m + 1][n] = 0;
pos[m + 2][n] = 0;
pos[m][n + 1] = 0;
pos[m + 1][n + 1] = 1;
pos[m + 1][n + 2] = 0;
pos[m][n + 2] = 0;
pos[m + 2][n + 1] = 0;
pos[m + 2][n + 2] = 1;
}
else if (c == '/') {
pos[m][n] = 0;
pos[m][n + 1] = 0;
pos[m][n + 2] = 1;
pos[m + 1][n] = 0;
pos[m + 2][n] = 1;
pos[m][n + 1] = 0;
pos[m + 1][n + 1] = 1;
pos[m + 2][n + 2] = 0;
pos[m + 1][n + 2] = 0;
}
}
}
row = 3 * row;
col = 3 * col;
for (int i = 0; i < row; i++)
for (int j = 0; j < col; j++)
if (pos[i][j] == 0) {
flag = 1;
step = 1;
dfs(i, j);
if (flag == 1) {
count++;
if (step > max)
max = step;
}
}
printf("Maze #%d:\n", t++);
if (count == 0)
printf("There are no cycles.\n\n");
else
printf("%d Cycles; the longest has length %d.\n\n", count, max / 3);
}
return 0;
}
