Algorithm
Problem link- https://www.spoj.com/problems/MAKEMAZE/
MAKEMAZE - VALIDATE THE MAZE
There are many algorithms to generate maze. (http://en.wikipedia.org/wiki/Maze_generation_algorithm). After generating the maze we’ve to validate whether it’s a valid maze or not. A valid maze has exactly one entry point and exactly one exit point (exactly 2 openings in the edges) and there must be at least one path from the entry point to exit point.
Given a maze, just find whether the maze is "valid" or "invalid".
Input
The first line consists of an integer t, the number of test cases. Then for each test case, the first line consists of two integers m and n, the number of rows and columns in the maze. Then contains the description of the matrix M of order mxn. M[i][j]=# represents a wall and M[i][j]='.' represents a space.
Output
For each test case find whether the maze is "valid" or "invalid".
Constraints
1<=t<=10000
1<=m<=20
1<=n<=20
Example
Input: 6 4 4 #### #... #.## #.## 5 5 #.### #..## ##..# #.#.# ###.# 1 1 . 5 1 # # . . # 2 2 #. .# 3 4 #..# #.## #.## Output: valid valid invalid valid invalid invalid
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <iostream>
#include <sstream>
#include <fstream>
#include <cassert>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <string>
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
using namespace std;
typedef pair<int,int> ii;
typedef vector<vector<pair<int,ii>> > vviii;
#define se second
#define fi first
#define pb push_back
const int INF = 0x3f3f3f3f;
vector<string> grid;
vector<string> visited;
int dx[] = {-1,1,0,0};
int dy[] = {0,0,1,-1};
int m,n,count,ecount;
void dfsUtil(int starti,int startj){
stack<ii> st;
st.push(ii(starti,startj));
visited[starti][startj]='T';
while(!st.empty()){
ii top = st.top();
st.pop();
for(int i=0;i<4;++i){
int cx = top.fi+dx[i];
int cy = top.se+dy[i];
if(cx<0||cx>m-1||cy<0||cy>n-1||grid[cx][cy]=='#'||visited[cx][cy]=='T')
continue;
if(cx==0||cy==0||cx==m-1||cy==n-1){
count++;
}
st.push(ii(cx,cy));
visited[cx][cy]='T';
}
}
}
void dfs(){
for(int i=0;i<m;++i){
for(int j=0;j<n;++j){
if(grid[i][j]=='.'&&(i==0||j==0||i==(m-1)||j==(n-1))){
ecount++;
dfsUtil(i,j);
if(count>1){
return ;
}
// reset visited if no path found
else if(count==0){
visited = vector<string> (m,string(n,'F'));
}
}
}
}
}
int main(){
int t;
scanf("%d",&t);
while(t--){
count=0;
//count the number of openings on the edges
ecount=0;
scanf("%d %d",&m,&n);
grid = vector<string> (m,string());
visited = vector<string> (m,string(n,'F'));
string input;
for(int i=0;i<m;++i){
cin>>input;
grid[i]=input;
}
dfs();
if(count>1||count==0||ecount>2){
printf("invalid\n");
}
else{
printf("valid\n");
}
}
return 0;
}Input
4 4
####
#...
#.##
#.##
5 5
#.###
#..##
##..#
#.#.#
###.#
1 1
.
5 1
#
#
.
.
#
2 2
#.
.#
3 4
#..#
#.##
#.##
Output
valid
invalid
valid
invalid
invalid
Demonstration
SPOJ Solution-MAKEMAZE VALIDATE THE MAZE-Solution in C, C++, Java, Python