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C++ Program to find out the number of jumps needed for a robot to reach a particular cell in a grid
Suppose, we have a grid of dimensions h x w. The grid is represented in a 2D array called ‘initGrid’, where each cell in the grid is either represented by a '#' or a '.'. '#' means that the grid contains an obstacle and '.' means that there is a path through that cell. Now, a robot is placed on a cell 'c' on the grid having row number x and column number y. The robot has to travel to another cell 'd' having row number p and column number q. Both the cell coordinates c and d are presented to us as integer pairs. Now, the robot can move from one cell to another in the following ways −
The robot can just walk from one cell to another if the cell it wants to move to is located vertically or horizontally adjacent to the cell it is currently in.
The robot can jump to any cell in the 5×5 area which has the center at the cell it is currently located.
The robot can only move to another cell in the grid if the destination cell does not contain an obstacle. The robot also cannot leave the grid.
We have to find out the number of jumps it will need to reach the destination.
So, if the input is like h = 4, w = 4, c = {2, 1}, d = {4, 4}, initGrid = {"#...", ".##.", "...#", "..#."}, then the output will be 1. The robot will need only one jump to reach its destination.
To solve this, we will follow these steps −
N:= 100
Define intger pairs s and t.
Define an array grid of size: N.
Define an array dst of size: N x N.
Define a struct node that contains integer values a, b, and e.
Define a function check(), this will take a, b,
return a >= 0 AND a < h AND b >= 0 AND b < w
Define a function bfs(), this will take a, b,
for initialize i := 0, when i < h, update (increase i by 1), do:
for initialize j := 0, when j < w, update (increase j by 1), do:
dst[i, j] := infinity
dst[a, b] := 0
Define one deque doubleq
Insert a node containing values {a, b, and dst[a, b]} at the end of doubleq
while (not doubleq is empty), do:
nd := first element of doubleq
if e value of nd > dst[a value of nd, b value of nd], then:
Ignore the following part, skip to the next iteration
for initialize diffx := -2, when diffx <= 2, update (increase diffx by 1), do:
for initialize diffy := -2, when diffy <= 2, update (increase diffy by 1), do:
tm := |diffx + |diffy||
nx := a value of nd + diffx, ny = b value of nd + diffy
if check(nx, ny) and grid[nx, ny] is same as '.', then:
w := (if tm > 1, then 1, otherwise 0)
if dst[a value of nd, b value of nd] + w < dst[nx, ny], then:
dst[nx, ny] := dst[a value of nd, b value of nd] + w
if w is same as 0, then:
insert node containing values ({nx, ny, dst[nx, ny]}) at the beginning of doubleq.
Otherwise
insert node containing values ({nx, ny, dst[nx, ny]}) at the end of doubleq.
s := c
t := d
(decrease first value of s by 1)
(decrease second value of s by 1)
(decrease first value of t by 1)
(decrease second value of t by 1)
for initialize i := 0, when i < h, update (increase i by 1), do:
grid[i] := initGrid[i]
bfs(first value of s, second value of s)
print(if dst[first value of t, second value of t] is same as infinity, then -1, otherwise dst[first value of t, second value of t])Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
#define N 100
int h, w;
pair<int, int> s, t;
string grid[N];
int dst[N][N];
struct node {
int a, b, e;
};
bool check(int a, int b) {
return a >= 0 && a < h && b >= 0 && b < w;
}
void bfs(int a, int b) {
for (int i = 0; i < h; i++) {
for (int j = 0; j < w; j++)
dst[i][j] = INF;
}
dst[a][b] = 0;
deque<node> doubleq;
doubleq.push_back({a, b, dst[a][b]});
while (!doubleq.empty()) {
node nd = doubleq.front();
doubleq.pop_front();
if (nd.e > dst[nd.a][nd.b])
continue;
for (int diffx = -2; diffx <= 2; diffx++) {
for (int diffy = -2; diffy <= 2; diffy++) {
int tm = abs(diffx) + abs(diffy);
int nx = nd.a + diffx, ny = nd.b + diffy;
if (check(nx, ny) && grid[nx][ny] == '.') {
int w = (tm > 1) ? 1 : 0;
if (dst[nd.a][nd.b] + w < dst[nx][ny]) {
dst[nx][ny] = dst[nd.a][nd.b] + w;
if (w == 0)
doubleq.push_front({nx, ny, dst[nx][ny]});
else
doubleq.push_back({nx, ny, dst[nx][ny]});
}
}
}
}
}
}
void solve(pair<int,int> c, pair<int, int> d, string initGrid[]){
s = c;
t = d;
s.first--, s.second--, t.first--, t.second--;
for(int i = 0; i < h; i++)
grid[i] = initGrid[i];
bfs(s.first, s.second);
cout << (dst[t.first][t.second] == INF ? -1 :
dst[t.first][t.second]) << '\n';
}
int main() {
h = 4, w = 4;
pair<int,int> c = {2, 1}, d = {4, 4};
string initGrid[] = {"#...", ".##.", "...#", "..#."};
solve(c, d, initGrid);
return 0;
}Input
4, 4, {2, 1}, {4, 4}, {"#...", ".##.", "...#", "..#."}Output
1
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