Given a matrix of 0 and 1 (0 is free space, 1 is wall). Find the shortest path from one cell to another, passing only through 0 and also without touching 1.
enter image description here
How can I do this using Lee's Algorithm?
class Solution {
public:
int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
// edge case: start or end not accessible
if (grid[0][0] || grid.back().back()) return -1;
// support variables
int res = 2, len = 1, maxX = grid[0].size() - 1, maxY = grid.size() - 1;
queue<pair<int, int>> q;
// edge case: single cell matrix
if (!maxX && !maxY) return 1 - (grid[0][0] << 1);
// adding the starting point
q.push({0, 0});
// marking start as visited
grid[0][0] = -1;
while (len) {
while (len--) {
// reading and popping the coordinates on the front of the queue
auto [cx, cy] = q.front();
q.pop();
for (int x = max(0, cx - 1), lmtX = min(cx + 1, maxX); x <= lmtX; x++) {
for (int y = max(0, cy - 1), lmtY = min(cy + 1, maxY); y <= lmtY; y++) {
// check if we reached the target
if (x == maxX && y == maxY) return res;
// marking it as visited and adding it to the q if it was still a valid cell
if (!grid[y][x]) {
grid[y][x] = -1;
q.push({x, y});
}
}
}
}
// preparing for the next loop
res++;
len = q.size();
}
return -1;
}
};
You start at top-left cell of a given grid. Some cells have wall, some
you can walk, and some cells have apple. You are given a time limit =
T, and you should reach bottom right cell by atmost T time. Find
maximum number of apples you can collect. You cannot visit a cell
twice. N, M, T <= 14.
I tried a lot of ideas, most promising one is this - rephrase problem as find shortest time to reach destination collecting atleast X apples. Then we could binary search on number of apples.
But I am not able to pin down a solution from last 6hours.
"You cannot visit a cell twice." this is causing me problem.
Any other idea or hint is appreciated.
Have you tried backtracking? Something like this?
// Heuristic: Manhattan distance to end
function dist(y, x, n, m){
return n - y + m - x - 2;
}
function getNext(i, j, n, m){
const ways = [];
if (i + 1 < n)
ways.push([i+1, j]);
if (i > 0)
ways.push([i-1, j]);
if (j + 1 < m)
ways.push([i, j+1]);
if (j > 0)
ways.push([i, j-1]);
return ways;
}
function f(M, T){
const WALL = 2;
const n = M.length;
const m = M[0].length;
const visited = new Array(n);
for (let i=0; i<n; i++)
visited[i] = new Array(m).fill(0);
let best = 0;
function backtrack(i, j, t, k){
if (i == n-1 && j == m-1){
best = Math.max(best, k + M[i][j]);
return;
}
for (const [ii, jj] of getNext(i, j, n, m)){
if (!visited[ii][jj] &&
M[ii][jj] != WALL &&
t + dist(ii, jj, n, m) <= T){
visited[ii][jj] = 1;
backtrack(ii, jj, t + 1, k + M[i][j]);
visited[ii][jj] = 0
}
}
}
backtrack(0, 0, 0, 0);
return best;
}
var N = 8;
var M = 8;
var T = 14;
var matrix = new Array(N);
for (let i=0; i<N; i++)
matrix[i] = new Array(M).fill(0);
// Apples
matrix[5][5] = 1;
matrix[5][6] = 1;
// Walls
matrix[5][7] = 2;
matrix[5][4] = 2;
matrix[4][4] = 2;
matrix[4][5] = 2;
console.log(f(matrix, T));
matrix[5][4] = 0;
console.log(f(matrix, T));
Given the constraints, you can use a simple recursive function to complete the problem.
Let solve(i,j,steps,vis) be the function, where (i,j) are current coordinates, time is the time remaining, and vis is the set of currently visited nodes. The answer will be solve(0,0,T,[]).
The simple recursion would be (using pseudo-code):
def solve(i,j,t,vis):
if (i<0 or i>=n or j<0 or j>=m) return -1
if ((i,j) in vis) return -1
if (cell[i][j] == WALL) return -1
if (t==0){
if (i==n-1 and j==m-1) return cell[i][j]
else return -1
}
if (i==n-1 and j==m-1) return cell[i][j]
max_here = cell[i][j]
temp = max(solve(i,j+1,t-1,vis+(i,j)), solve(i,j-1,t-1,vis+(i,j)), solve(i+1,j,t-
1,vis+(i,j)), solve(i-1,j,t-1,vis+(i,j))) #assuming movement in 4 directions
if (temp==-1) return -1 # since none of the neighbours lead to destination
return max_here+temp
I participated in a programming competition at my University. I solved all the questions except this one. Now I am practicing this question to improve my skills. But I can't figure out the algorithm. If there is any algorithm existing please update me. Or any similar algorithm is present then please tell me I will change it according to this question.
This is what I want to do.
The First line of input is the distance between two points.
After that, each subsequent line contains a pair of numbers indicating the length of cable and quantity of that cable. These cables are used to join the two points.
Input is terminated by 0 0
Output:
The output should contain a single integer representing the minimum number of joints possible to build the requested length of cableway. If no solution possible than print "No solution".
Sample Input
444
16 2
3 2
2 2
30 3
50 10
45 12
8 12
0 0
Sample Output
10
Thanks guys. I found a solution from "Perfect subset Sum" problem and then made a few changes in it. Here's the code.
#include <bits/stdc++.h>
using namespace std;
bool dp[100][100];
int sizeOfJoints = -1;
void display(const vector<int>& v)
{
if (sizeOfJoints == -1)
{
sizeOfJoints = v.size() - 1;
}
else if (v.size()< sizeOfJoints)
{
sizeOfJoints = v.size() - 1;
}
}
// A recursive function to print all subsets with the
// help of dp[][]. Vector p[] stores current subset.
void printSubsetsRec(int arr[], int i, int sum, vector<int>& p)
{
// If sum becomes 0
if (sum == 0)
{
display(p);
return;
}
if(i<=0 || sum<0)
return;
// If given sum can be achieved after ignoring
// current element.
if (dp[i-1][sum])
{
// Create a new vector to store path
//vector<int> b = p;
printSubsetsRec(arr, i-1, sum, p);
}
// If given sum can be achieved after considering
// current element.
if (sum >= arr[i-1] && dp[i-1][sum-arr[i-1]])
{
p.push_back(arr[i-1]);
printSubsetsRec(arr, i-1, sum-arr[i-1], p);
p.pop_back();
}
}
// all subsets of arr[0..n-1] with sum 0.
void printAllSubsets(int arr[], int n, int sum)
{
if (n == 0 || sum < 0)
return;
// If sum is 0, then answer is true
for (int i = 0; i <= n; i++)
dp[i][0] = true;
// If sum is not 0 and set is empty, then answer is false
for (int i = 1; i <= sum; i++)
dp[0][i] = false;
// Fill the subset table in botton up manner
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= sum; j++)
{
if(j<arr[i-1])
dp[i][j] = dp[i-1][j];
if (j >= arr[i-1])
dp[i][j] = dp[i-1][j] ||
dp[i - 1][j-arr[i-1]];
}
}
if (dp[n][sum] == false)
{
return;
}
// Now recursively traverse dp[][] to find all
// paths from dp[n-1][sum]
vector<int> p;
printSubsetsRec(arr, n, sum, p);
}
// Driver code
int main()
{
int input[2000];
int inputIndex = 0;
int i = 0;
int distance = 0;
cout<< "Enter Input: " <<endl;
cin>> distance;
while(true)
{
int temp1 = 0;
int temp2 = 0;
cin>> temp1;
cin>> temp2;
if (temp1 == 0 && temp2 == 0)
{
break;
}
for (i = 0; i < temp2; i++)
input[inputIndex++] = temp1;
}
cout<< "Processing output. Please wait: " <<endl;
printAllSubsets(input, inputIndex, distance);
if(sizeOfJoints != -1)
cout<<sizeOfJoints;
else
cout<<"No Solution Possible";
return 0;
}
I am trying to solve the n-queen problem (placing n queens on a nxn board without any two queens attacking each other) by defining a function that takes an nxn boolean array of falses and should fill the answer with true where the queens should be. I am getting incorrect answer but can't see why the recursion doesn't work!
bool check(bool ** board, int n, int row, int col){
if(row == 0) return true;
for(int r = 0 ; r < row ; ++r){
if(board[r][col]) return false;
int left = max(col - row + r, 0), right = min(col + row - r, n-1);
if(board[r][left] || board[r][right]) return false;
}
return true;
}
bool queen(bool ** board, int n, int level = 0 ){
for(int col = 0 ; col < n ; ++col){
if( !check(board, n, level, col) ) continue;
board[ level ][ col ] = true;
if( level == n-1 ) return true;
if( queen(board, n, level+1) ) return true;
board[ level ][ col ] = false;
}
return false;
}
in main() i dynamically create bool ** board, and fill it with false, then call queen(board, n).
The weird thing is that it is giving the correct solution except for n=4,6!
Any help is greatly appreciated!
Your fault is the min/max.Operation, so you don't check straight lines.
This should do the trick:
int left = col - row + r;
int right = col + row - r;
if ( left >= 0 && board[r][left] || right < n && board[r][right])
return false;
I have asked a similar question some days ago, but I have yet to find an efficient way of solving my problem.
I'm developing a simple console game, and I have a 2D array like this:
1,0,0,0,1
1,1,0,1,1
0,1,0,0,1
1,1,1,1,0
0,0,0,1,0
I am trying to find all the areas that consist of neighboring 1's (4-way connectivity). So, in this example the 2 areas are as following:
1
1,1
1
1,1,1,1
1
and :
1
1,1
1
The algorithm, that I've been working on, finds all the neighbors of the neighbors of a cell and works perfectly fine on this kind of matrices. However, when I use bigger arrays (like 90*90) the program is very slow and sometimes the huge arrays that are used cause stack overflows.
One guy on my other question told me about connected-component labelling as an efficient solution to my problem.
Can somebody show me any C++ code which uses this algorithm, because I'm kinda confused about how it actually works along with this disjoint-set data structure thing...
Thanks a lot for your help and time.
I'll first give you the code and then explain it a bit:
// direction vectors
const int dx[] = {+1, 0, -1, 0};
const int dy[] = {0, +1, 0, -1};
// matrix dimensions
int row_count;
int col_count;
// the input matrix
int m[MAX][MAX];
// the labels, 0 means unlabeled
int label[MAX][MAX];
void dfs(int x, int y, int current_label) {
if (x < 0 || x == row_count) return; // out of bounds
if (y < 0 || y == col_count) return; // out of bounds
if (label[x][y] || !m[x][y]) return; // already labeled or not marked with 1 in m
// mark the current cell
label[x][y] = current_label;
// recursively mark the neighbors
for (int direction = 0; direction < 4; ++direction)
dfs(x + dx[direction], y + dy[direction], current_label);
}
void find_components() {
int component = 0;
for (int i = 0; i < row_count; ++i)
for (int j = 0; j < col_count; ++j)
if (!label[i][j] && m[i][j]) dfs(i, j, ++component);
}
This is a common way of solving this problem.
The direction vectors are just a nice way to find the neighboring cells (in each of the four directions).
The dfs function performs a depth-first-search of the grid. That simply means it will visit all the cells reachable from the starting cell. Each cell will be marked with current_label
The find_components function goes through all the cells of the grid and starts a component labeling if it finds an unlabeled cell (marked with 1).
This can also be done iteratively using a stack.
If you replace the stack with a queue, you obtain the bfs or breadth-first-search.
This can be solved with union find (although DFS, as shown in the other answer, is probably a bit simpler).
The basic idea behind this data structure is to repeatedly merge elements in the same component. This is done by representing each component as a tree (with nodes keeping track of their own parent, instead of the other way around), you can check whether 2 elements are in the same component by traversing to the root node and you can merge nodes by simply making the one root the parent of the other root.
A short code sample demonstrating this:
const int w = 5, h = 5;
int input[w][h] = {{1,0,0,0,1},
{1,1,0,1,1},
{0,1,0,0,1},
{1,1,1,1,0},
{0,0,0,1,0}};
int component[w*h];
void doUnion(int a, int b)
{
// get the root component of a and b, and set the one's parent to the other
while (component[a] != a)
a = component[a];
while (component[b] != b)
b = component[b];
component[b] = a;
}
void unionCoords(int x, int y, int x2, int y2)
{
if (y2 < h && x2 < w && input[x][y] && input[x2][y2])
doUnion(x*h + y, x2*h + y2);
}
int main()
{
for (int i = 0; i < w*h; i++)
component[i] = i;
for (int x = 0; x < w; x++)
for (int y = 0; y < h; y++)
{
unionCoords(x, y, x+1, y);
unionCoords(x, y, x, y+1);
}
// print the array
for (int x = 0; x < w; x++)
{
for (int y = 0; y < h; y++)
{
if (input[x][y] == 0)
{
cout << ' ';
continue;
}
int c = x*h + y;
while (component[c] != c) c = component[c];
cout << (char)('a'+c);
}
cout << "\n";
}
}
Live demo.
The above will show each group of ones using a different letter of the alphabet.
p i
pp ii
p i
pppp
p
It should be easy to modify this to get the components separately or get a list of elements corresponding to each component. One idea is to replace cout << (char)('a'+c); above with componentMap[c].add(Point(x,y)) with componentMap being a map<int, list<Point>> - each entry in this map will then correspond to a component and give a list of points.
There are various optimisations to improve the efficiency of union find, the above is just a basic implementation.
You could also try this transitive closure approach, however the triple loop for the transitive closure slows things up when there are many separated objects in the image, suggested code changes welcome
Cheers
Dave
void CC(unsigned char* pBinImage, unsigned char* pOutImage, int width, int height, int CON8)
{
int i, j, x, y, k, maxIndX, maxIndY, sum, ct, newLabel=1, count, maxVal=0, sumVal=0, maxEQ=10000;
int *eq=NULL, list[4];
int bAdd;
memcpy(pOutImage, pBinImage, width*height*sizeof(unsigned char));
unsigned char* equivalences=(unsigned char*) calloc(sizeof(unsigned char), maxEQ*maxEQ);
// modify labels this should be done with iterators to modify elements
// current column
for(j=0; j<height; j++)
{
// current row
for(i=0; i<width; i++)
{
if(pOutImage[i+j*width]>0)
{
count=0;
// go through blocks
list[0]=0;
list[1]=0;
list[2]=0;
list[3]=0;
if(j>0)
{
if((i>0))
{
if((pOutImage[(i-1)+(j-1)*width]>0) && (CON8 > 0))
list[count++]=pOutImage[(i-1)+(j-1)*width];
}
if(pOutImage[i+(j-1)*width]>0)
{
for(x=0, bAdd=true; x<count; x++)
{
if(pOutImage[i+(j-1)*width]==list[x])
bAdd=false;
}
if(bAdd)
list[count++]=pOutImage[i+(j-1)*width];
}
if(i<width-1)
{
if((pOutImage[(i+1)+(j-1)*width]>0) && (CON8 > 0))
{
for(x=0, bAdd=true; x<count; x++)
{
if(pOutImage[(i+1)+(j-1)*width]==list[x])
bAdd=false;
}
if(bAdd)
list[count++]=pOutImage[(i+1)+(j-1)*width];
}
}
}
if(i>0)
{
if(pOutImage[(i-1)+j*width]>0)
{
for(x=0, bAdd=true; x<count; x++)
{
if(pOutImage[(i-1)+j*width]==list[x])
bAdd=false;
}
if(bAdd)
list[count++]=pOutImage[(i-1)+j*width];
}
}
// has a neighbour label
if(count==0)
pOutImage[i+j*width]=newLabel++;
else
{
pOutImage[i+j*width]=list[0];
if(count>1)
{
// store equivalences in table
for(x=0; x<count; x++)
for(y=0; y<count; y++)
equivalences[list[x]+list[y]*maxEQ]=1;
}
}
}
}
}
// floyd-Warshall algorithm - transitive closure - slow though :-(
for(i=0; i<newLabel; i++)
for(j=0; j<newLabel; j++)
{
if(equivalences[i+j*maxEQ]>0)
{
for(k=0; k<newLabel; k++)
{
equivalences[k+j*maxEQ]= equivalences[k+j*maxEQ] || equivalences[k+i*maxEQ];
}
}
}
eq=(int*) calloc(sizeof(int), newLabel);
for(i=0; i<newLabel; i++)
for(j=0; j<newLabel; j++)
{
if(equivalences[i+j*maxEQ]>0)
{
eq[i]=j;
break;
}
}
free(equivalences);
// label image with equivalents
for(i=0; i<width*height; i++)
{
if(pOutImage[i]>0&&eq[pOutImage[i]]>0)
pOutImage[i]=eq[pOutImage[i]];
}
free(eq);
}
very useful Document => https://docs.google.com/file/d/0B8gQ5d6E54ZDM204VFVxMkNtYjg/edit
java application - open source - extract objects from image - connected componen labeling => https://drive.google.com/file/d/0B8gQ5d6E54ZDTVdsWE1ic2lpaHM/edit?usp=sharing
import java.util.ArrayList;
public class cclabeling
{
int neighbourindex;ArrayList<Integer> Temp;
ArrayList<ArrayList<Integer>> cc=new ArrayList<>();
public int[][][] cclabel(boolean[] Main,int w){
/* this method return array of arrays "xycc" each array contains
the x,y coordinates of pixels of one connected component
– Main => binary array of image
– w => width of image */
long start=System.nanoTime();
int len=Main.length;int id=0;
int[] dir={-w-1,-w,-w+1,-1,+1,+w-1,+w,+w+1};
for(int i=0;i<len;i+=1){
if(Main[i]){
Temp=new ArrayList<>();
Temp.add(i);
for(int x=0;x<Temp.size();x+=1){
id=Temp.get(x);
for(int u=0;u<8;u+=1){
neighbourindex=id+dir[u];
if(Main[neighbourindex]){
Temp.add(neighbourindex);
Main[neighbourindex]=false;
}
}
Main[id]=false;
}
cc.add(Temp);
}
}
int[][][] xycc=new int[cc.size()][][];
int x;int y;
for(int i=0;i<cc.size();i+=1){
xycc[i]=new int[cc.get(i).size()][2];
for(int v=0;v<cc.get(i).size();v+=1){
y=Math.round(cc.get(i).get(v)/w);
x=cc.get(i).get(v)-y*w;
xycc[i][v][0]=x;
xycc[i][v][1]=y;
}
}
long end=System.nanoTime();
long time=end-start;
System.out.println("Connected Component Labeling Time =>"+time/1000000+" milliseconds");
System.out.println("Number Of Shapes => "+xycc.length);
return xycc;
}
}
Please find below the sample code for connected component labeling . The code is written in JAVA
package addressextraction;
public class ConnectedComponentLabelling {
int[] dx={+1, 0, -1, 0};
int[] dy={0, +1, 0, -1};
int row_count=0;
int col_count=0;
int[][] m;
int[][] label;
public ConnectedComponentLabelling(int row_count,int col_count) {
this.row_count=row_count;
this.col_count=col_count;
m=new int[row_count][col_count];
label=new int[row_count][col_count];
}
void dfs(int x, int y, int current_label) {
if (x < 0 || x == row_count) return; // out of bounds
if (y < 0 || y == col_count) return; // out of bounds
if (label[x][y]!=0 || m[x][y]!=1) return; // already labeled or not marked with 1 in m
// mark the current cell
label[x][y] = current_label;
// System.out.println("****************************");
// recursively mark the neighbors
int direction = 0;
for (direction = 0; direction < 4; ++direction)
dfs(x + dx[direction], y + dy[direction], current_label);
}
void find_components() {
int component = 0;
for (int i = 0; i < row_count; ++i)
for (int j = 0; j < col_count; ++j)
if (label[i][j]==0 && m[i][j]==1) dfs(i, j, ++component);
}
public static void main(String[] args) {
ConnectedComponentLabelling l=new ConnectedComponentLabelling(4,4);
l.m[0][0]=0;
l.m[0][1]=0;
l.m[0][2]=0;
l.m[0][3]=0;
l.m[1][0]=0;
l.m[1][1]=1;
l.m[1][2]=0;
l.m[1][3]=0;
l.m[2][0]=0;
l.m[2][1]=0;
l.m[2][2]=0;
l.m[2][3]=0;
l.m[3][0]=0;
l.m[3][1]=1;
l.m[3][2]=0;
l.m[3][3]=0;
l.find_components();
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 4; j++) {
System.out.print(l.label[i][j]);
}
System.out.println("");
}
}
}