KnightL on a Chessboard
Problem Statement :
KnightL is a chess piece that moves in an L shape. We define the possible moves of as any movement from some position to some satisfying either of the following: Note that and allow for the same exact set of movements. For example, the diagram below depicts the possible locations that or can move to from its current location at the center of a chessboard: Observe that for each possible movement, the Knight moves units in one direction (i.e., horizontal or vertical) and unit in the perpendicular direction. Input Format A single integer denoting . Constraints Output Format Print exactly lines of output in which each line (where ) contains space-separated integers describing the minimum number of moves must make for each respective (where ). If some cannot reach position , print -1 instead.
Solution :
Solution in C :
In C++ :
#include<bits/stdc++.h>
using namespace std;
const int BIG = 2e9;
int n, d[50][50];
void can(queue<pair<int, int>> &q, int i, int j, int dd){
if(i >= 0 && i < n && j >= 0 && j < n && d[i][j] > dd){
d[i][j] = dd;
q.push({i, j});
}
}
int bfs(int n, int a, int b){
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
d[i][j] = BIG;
}
}
d[0][0] = 0;
queue<pair<int, int>> q;
q.push({0, 0});
while(!q.empty()){
pair<int, int> v = q.front();
q.pop();
int x = v.first;
int y = v.second;
int dist = d[x][y] + 1;
can(q, x + a, y + b, dist);
can(q, x + a, y - b, dist);
can(q, x - a, y + b, dist);
can(q, x - a, y - b, dist);
can(q, x + b, y + a, dist);
can(q, x + b, y - a, dist);
can(q, x - b, y + a, dist);
can(q, x - b, y - a, dist);
}
int ans = d[n - 1][n - 1];
if(ans == BIG){
return -1;
}
return ans;
}
int main(){
cin >> n;
for(int i = 1; i < n; i++){
for(int j = 1; j < n; j++){
cout << bfs(n, i, j) << " ";
}
cout << "\n";
}
}
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
// your code goes here
int[][] results = new int[n-1][n-1];
for(int i = 1; i < n; i++){ //decres i and j when entering
for(int j = 1; j < n; j++){
int[][] T = new int[n][n];
Queue<Tupel> q = new LinkedList<Tupel>();
q.add(new Tupel(0,0));
while(!q.isEmpty()){
Tupel t = q.remove();
if(t.x == 0 && t.y == 0){
T[0][0] = 1;
}
if(t.x-i >= 0 && t.y-j >= 0 && T[t.x-i][t.y-j] == 0){
T[t.x-i][t.y-j] = T[t.x][t.y]+1;
q.add(new Tupel(t.x-i, t.y-j));
}
if(t.x-j >= 0 && t.y-i >= 0 && T[t.x-j][t.y-i] == 0){
T[t.x-j][t.y-i] = T[t.x][t.y]+1;
q.add(new Tupel(t.x-j, t.y-i));
}
if(t.x-i >= 0 && t.y+j < n && T[t.x-i][t.y+j] == 0){
T[t.x-i][t.y+j] = T[t.x][t.y]+1;
q.add(new Tupel(t.x-i, t.y+j));
}
if(t.x-j >= 0 && t.y+i < n && T[t.x-j][t.y+i] == 0){
T[t.x-j][t.y+i] = T[t.x][t.y]+1;
q.add(new Tupel(t.x-j, t.y+i));
}
if(t.x+i < n && t.y+j < n && T[t.x+i][t.y+j] == 0){
T[t.x+i][t.y+j] = T[t.x][t.y]+1;
if(t.x+i == n-1 && t.y+j == n-1){
break;
}
q.add(new Tupel(t.x+i, t.y+j));
}
if(t.x+j < n && t.y+i < n && T[t.x+j][t.y+i] == 0){
T[t.x+j][t.y+i] = T[t.x][t.y]+1;
if(t.x+i == n-1 && t.y+j == n-1){
break;
}
q.add(new Tupel(t.x+j, t.y+i));
}
if(t.x+i < n && t.y-j >= 0 && T[t.x+i][t.y-j] == 0){
T[t.x+i][t.y-j] = T[t.x][t.y]+1;
q.add(new Tupel(t.x+i, t.y-j));
}
if(t.x+j < n && t.y-i >= 0 && T[t.x+j][t.y-i] == 0){
T[t.x+j][t.y-i] = T[t.x][t.y]+1;
q.add(new Tupel(t.x+j, t.y-i));
}
}
results[i-1][j-1] = T[n-1][n-1]-1;
}
}
for(int i = 0; i < n-1; i++){
for(int j = 0; j < n-1; j++){
System.out.print(results[i][j] + " ");
}
System.out.println();
}
}
}
class Tupel{
int x;
int y;
public Tupel(int x, int y){
this.x = x;
this.y = y;
}
}
In C :
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int n;
int **checked;
typedef struct node{
int i;
int j;
int g;
int h;
struct node *next;
}node_t;
node_t *head = NULL;
node_t *tail = NULL;
void insert(int i, int j, int g, int h){
if(checked[i][j]==1) return;
//printf("Insert: %d %d\n",i,j);
checked[i][j] = 1;
node_t *newnode = (node_t*)malloc(sizeof(node_t));
newnode->i = i;
newnode->j = j;
newnode->g = g;
newnode->h = h;
newnode->next = NULL;
if(head==NULL){
head=newnode;
}
else{
tail->next=newnode;
}
tail=newnode;
}
int isEmpty(){
if(head==NULL){
return 1;
}
else return 0;
}
node_t *dequeue(){
node_t *ret;
ret = head;
head = head->next;
return ret;
}
void testone(int i, int j,int g){
if(i<0) return;
if(i>=n) return;
if(j<0) return;
if(j>=n) return;
insert(i,j,g,0);
}
void initchecked(){
int i,j;
for(i=0; i<n; i++){
for(j=0; j<n; j++){
checked[i][j] = 0;
}
}
}
void Astro(int a, int b){
node_t *node;
tail = NULL;
head = NULL;
initchecked();
insert(0,0,0,0);
while(!isEmpty()){
node = dequeue();
//printf("Test: %d %d\n",node->i, node->j);
if(node==NULL){
printf("-1 ");
return ;
}
else if(node->i==n-1 && node->j==n-1){
printf("%d ",node->g);
return ;
}
testone(node->i - b, node->j + a, node->g+1);
testone(node->i - a, node->j + b, node->g+1);
testone(node->i + a, node->j + b, node->g+1);
testone(node->i + b, node->j + a, node->g+1);
testone(node->i - b, node->j - a, node->g+1);
testone(node->i - a, node->j - b, node->g+1);
testone(node->i + a, node->j - b, node->g+1);
testone(node->i + b, node->j - a, node->g+1);
}
printf("-1 ");
}
int main(){
scanf("%d",&n);
checked = (int **)malloc(n*sizeof(int *));
for(int i=0; i<n; i++){
checked[i] = (int *)malloc(n*sizeof(int *));
}
for(int i=1; i<=n-1; i++){
for(int j=1; j<=n-1; j++){
Astro(i,j);
}
printf("\n");
}
return 0;
}
In Python3 :
#!/bin/python3
import sys
n = int(input().strip())
# your code goes here
A=[]
for a in range(1,n):
B=[]
for b in range(1,n):
P={(0,0)}
c=0
f=True
V=set()
while f:
c+=1
Q=set()
V=V|P
for p in P:
if (p[0]==n-1)&(p[1]==n-1):
f=False
break
if (0<=p[0]+a<n)&(0<=p[1]+b<n):
Q.add((p[0]+a,p[1]+b))
if (0<=p[0]-a<n)&(0<=p[1]-b<n):
Q.add((p[0]-a,p[1]-b))
if (0<=p[0]-a<n)&(0<=p[1]+b<n):
Q.add((p[0]-a,p[1]+b))
if (0<=p[0]+a<n)&(0<=p[1]-b<n):
Q.add((p[0]+a,p[1]-b))
if (0<=p[0]+b<n)&(0<=p[1]+a<n):
Q.add((p[0]+b,p[1]+a))
if (0<=p[0]-b<n)&(0<=p[1]-a<n):
Q.add((p[0]-b,p[1]-a))
if (0<=p[0]-b<n)&(0<=p[1]+a<n):
Q.add((p[0]-b,p[1]+a))
if (0<=p[0]+b<n)&(0<=p[1]-a<n):
Q.add((p[0]+b,p[1]-a))
P=set()
P=P|Q
P=P-V
if len(P)==0:
break
if not f:
B.append(c-1)
else:
B.append(-1)
A.append(B)
for a in A:
for b in a:
print(b,end= " ")
print("")
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