Sha 705 Slash Maze

Slash Maze

By filling a rectangle with slashes (/) and backslashes (), you can generate nice little mazes. Here is an example:

 

As you can see, paths in the maze cannot branch, so the whole maze only contains cyclic paths and paths entering somewhere and leaving somewhere else. we are only interested in the cycles. in our example, there are two of them.

Your task is to write a program that counts the cycles and finds the length of the longest one. the length is defined as the number of small squares the cycle consists of (the ones bordered by gray lines in the picture ). in this example, the long cycle has length 16 and the short one length 4.

Input
The input contains several maze descriptions. each description begins with one line containing two integers w and h (), the width and the height of the maze. the next h lines represent the maze itself, and contain w characters each; all these characters will be either ''/" or ''\".

The input is terminated by a test case beginning with w = h = 0. This case shocould not be processed.

Output
For each maze, first output the line ''maze # n: '', where n is the number of the Maze. then, output the line ''kcycles; the longest has length l. '', where k is the number of cycles in the maze and l the length of the longest of the cycles. if the maze does not contain any cycles, output the line ''there are no cycles. ".

Output a blank line after each test case.

Sample Input

6 4
\//\\/
\///\/
//\\/\
\/\///
3 3
///
\//
\\\
0 0

Sample Output

Maze #1:
2 Cycles; the longest has length 16.

Maze #2:
There are no cycles.

Question: Input '/' and '\' to form a maze .. Find out the number of loops that can be formed in the Maze and the longest Loop Length in the loop...

Idea: here we use two methods...

The first method is to convert each grid in the maze into a 3*3 grid .. The slashes are represented by 1, and the rest are represented by 0. After each slashes are saved. Search for a new map. In this way, you can search for BFS. If the value is 0 in the map, the border can be reached. It is impossible to form a loop.

If not. It is inevitable that a loop can be formed. After each search, mark the adjacent parts. You do not need to repeat the search. A loop can be formed if it is found. The length is divided by 3.

 

#include <stdio.h>#include <string.h>int n, m;int d[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};int map[275][275];int maxx;char sb;struct Q{    int x;    int y;} q[66666];int vis[275][275];int numc;void bfs(int x, int y){    memset(q, 0, sizeof(q));    memset(vis, 0, sizeof(vis));    int h = 0;    int r = 1;    int num = 1;    q[h].x = x;    q[h].y = y;    vis[x][y] = 1;    while (h < r)    {for (int i = 0; i < 4; i ++){    int xx = q[h].x + d[i][0];    int yy = q[h].y + d[i][1];    if (xx < 0 || xx >= 3 * n || yy < 0 || yy >=  3 * m)    {return;    }    if (map[xx][yy] == 0 && vis[xx][yy] == 0)    {vis[xx][yy] = 1;q[r].x = xx;q[r].y = yy;num ++;r ++;    }}h ++;    }    numc ++;    if (maxx < num / 3)maxx = num / 3;}void bfs2(int x, int y){    memset(q, 0, sizeof(q));    int h = 0;    int r = 1;    q[h].x = x;    q[h].y = y;    map[x][y] = 1;    while (h < r)    {for (int i = 0; i < 4; i ++){    int xx = q[h].x + d[i][0];    int yy = q[h].y + d[i][1];    if (map[xx][yy] == 0 && xx >= 0 && xx < 3 * n && yy >= 0 && yy < 3 * m)    {map[xx][yy] = 1;q[r].x = xx;q[r].y = yy;r ++;    }}h ++;    }}int main(){    int tt = 1;    while (scanf("%d%d", &m, &n) != EOF && n + m)    {maxx = 0;numc = 0;memset(map, 0, sizeof(map));getchar();for (int i = 0; i < n; i ++){    for (int j = 0; j < m; j ++)    {scanf("%c", &sb);if (sb == '\\'){    map[i * 3][j * 3] = 1;    map[i * 3 + 1][j * 3 + 1] = 1;    map[i * 3 + 2][j * 3 + 2] = 1;}if (sb == '/'){    map[i * 3][j * 3 + 2] = 1;    map[i * 3 + 1][j * 3 + 1] = 1;    map[i * 3 + 2][j * 3] = 1;}    }    getchar();}for (int i = 0; i < 3 * n; i ++){    for (int j = 0; j < 3 * m; j ++)    {if (map[i][j] == 0){    bfs(i, j);     bfs2(i, j);}    }}printf("Maze #%d:\n", tt ++);if (numc)    printf("%d Cycles; the longest has length %d.\n\n", numc, maxx);else    printf("There are no cycles.\n\n");    }    return 0;}

The second method is related to optics in physics .. In fact, we can regard each wall as a mirror, and the path as a light. The reflection path is unique after the light is injected into the mirror. Search by path every time .. If yes, the boundary is exceeded .. This path is not available. If the return start point is used, a loop exists .. This method is annoying to write. Consider that the wall is '\', '/', and the path's incident direction .. However, the time is much faster than the previous method.

 

#include <stdio.h>#include <string.h>int bo;int n, m;int xxx, yyy;int www;int numc;int d[4][2] = {{-1,0},{0,1},{1,0},{0,-1}};char map[80][80];int vis[2][80][80];int maxx;void dfs(int x, int y, int f, int w, int bu){    if (x < 1 || x > n || y < 1 || y > m)return;    if (bo == 0)bo = 1;    else    {if (x == xxx && y == yyy && w == www){    numc ++;    if (maxx < bu)maxx = bu;    return;}    }    vis[w][x][y] = 1;    int xx = x + d[f][0];    int yy = y + d[f][1];    if (f == 0)    {if (map[xx][yy] == '\\'){    dfs(xx, yy, 3, 1, bu + 1);}if (map[xx][yy] == '/'){    dfs(xx, yy, 1, 1, bu + 1);}    }    if (f == 1)    {if (map[xx][yy] == '\\'){    dfs(xx, yy, 2, 1, bu + 1);}if (map[xx][yy] == '/'){    dfs(xx, yy, 0, 0, bu + 1);}    }    if (f == 2)    {if (map[xx][yy] == '\\'){    dfs(xx, yy, 1, 0, bu + 1);}if (map[xx][yy] == '/'){    dfs(xx, yy, 3, 0, bu + 1);}    }    if (f == 3)    {if (map[xx][yy] == '\\'){    dfs(xx, yy, 0, 0, bu + 1);}if (map[xx][yy] == '/'){    dfs(xx, yy, 2, 1, bu + 1);}    }}int main(){    int tt = 1;    while (scanf("%d%d", &m, &n) != EOF && n + m)    {maxx = 0;numc = 0;getchar();memset(map, 0, sizeof(map));memset(vis, 0, sizeof(vis));for (int i = 1; i <= n; i ++){    gets(map[i] + 1);}for (int i = 1; i <= n; i ++)    for (int j = 1; j <= m; j ++)    {xxx = i; yyy = j;if (vis[0][i][j] == 0){    www = 0;    bo = 0;    dfs(i, j, 0, 0, 0);    if (map[i][j] == '/')    {bo = 0;dfs(i, j, 3, 0, 0);    }    if (map[i][j] == '\\')    {bo = 0;dfs(i, j, 1, 0, 0);    }}if (vis[1][i][j] == 0){    www = 1;    if (map[i][j] == '/')    {bo = 0;dfs(i, j, 1, 1, 0);    }    if (map[i][j] == '\\')    {bo = 0;dfs(i, j, 3, 1, 0);    }    bo = 0;    dfs(i, j, 2, 1, 0);}    }printf("Maze #%d:\n", tt ++);if (numc)    printf("%d Cycles; the longest has length %d.\n\n", numc / 2, maxx);else    printf("There are no cycles.\n\n");    }    return 0;}