C++ 程序,找出游遍所有给定坐标的费用
假设,我们得到了 n 个三维坐标。从坐标 (a, b, c) 到 (x, y, z) 的移动成本是 ∣ x − a∣ + ∣ y − b∣ + max(0, z − c)。我们从第一个坐标开始,然后至少访问所有坐标一次,然后返回第一个坐标。我们必须算出整个行程的总成本。这些坐标在数组 'coords' 中提供给了我们。
因此,如果输入是 n = 3,coords = {{1, 1, 0}, {1, 3, 4}, {3, 2, 2}},则输出将为 12。
为了解决这个问题,我们将按照以下步骤进行 −
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
示例
让我们看看以下实现,以获得更好的理解 −
#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; }
输入
3, {{1, 1, 0}, {1, 3, 4}, {3, 2, 2}}
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输出
12
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