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C++ Program to find out the cost to travel all the given coordinates
Suppose, we are given n three-dimensional coordinates. The cost to travel from coordinate (a, b, c) to (x, y, z) is ∣ x − a∣ + ∣ y − b∣ + max(0, z − c). We start from the first coordinate, then visit all the coordinates at least once, and then return to the first coordinate. We have to find out the total cost of this whole trip. The coordinates are given to us in the array 'coords'.
So, if the input is like n = 3, coords = {{1, 1, 0}, {1, 3, 4}, {3, 2, 2}}, then the output will be 12.
To solve this, we will follow these steps −
Define one 2D array tpa. tpa[1, 0] := 0 for initialize i := 1, when i < 2n, update (increase i by 1), do: for initialize j := 0, when j < n, update (increase j by 1), do: if i mod 2 is same as 0, then: Ignore following part, skip to the next iteration for initialize t := 0, when t < n, update (increase t by 1), do: x := first value of coords[t] y := second value of coords[t] z := third value of coords[t] p := first value of coords[j] q := second value of coords[j] r := third value of coords[j] tpa[i OR (1 bitwise left shift t)][t] := minimum of (tpa[i|(1 bitwise left shift t)][t], tpa[i][j] + |x - p| + |y - q| + maximum of (0, z - r)) res := infinity for initialize i := 0, when i < n, update (increase i by 1), do: x := first value of coords[0] y := second value of coords[0] z := third value of coords[0] p := first value of coords[i] q := second value of coords[i] r := third value of coords[i] res := minimum of (res and tpa[2n - 1, i] + |x - p| + |y - q| + maximum of (0 and z - r)) return res
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 solve(int n, vector<tuple<int,int,int>> coords){
vector<vector<int>> tpa(pow(2, n), vector<int>(n, INF));
tpa[1][0] = 0;
for(int i = 1; i < pow(2,n); i++) {
for(int j = 0; j < n; j++){
if(i % 2 == 0)
continue;
for(int t = 0; t < n; t++) {
int x, y, z, p, q, r;
tie(x, y, z) = coords[t];
tie(p, q, r) = coords[j];
tpa[i | (1 << t)][t] = min(tpa[i|(1 << t)][t], tpa[i][j] + abs(x - p) + abs(y - q) + max(0, z - r));
}
}
}
int res = INF;
for(int i = 0; i < n; i++) {
int x, y, z, p, q, r;
tie(x, y, z) = coords[0];
tie(p, q, r) = coords[i];
res = min(res, tpa[pow(2, n) - 1][i] + abs(x - p) + abs(y - q) + max(0, z - r));
}
return res;
}
int main() {
int n = 3;
vector<tuple<int,int,int>> coords = {{1, 1, 0}, {1, 3, 4}, {3, 2, 2}};
cout<< solve(n, coords);
return 0;
}Input
3, {{1, 1, 0}, {1, 3, 4}, {3, 2, 2}}Output
12
Updated on: 25-Feb-2022
183 Views
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