C++程序找出旅行所有给定坐标的成本
假设给定 n 个三维坐标。从坐标 (a, b, c) 到 (x, y, z) 的成本为 ∣ 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，然后返回到第一个坐标。我们必须找出整个行程的总费用。坐标在数组“coords”中提供给我们。
因此，如果输入类似于 n = 3，coords = {{1, 1, 0}, {1, 3, 4}, {3, 2, 2}}，那么输出将为 12。
为了解决这个问题，我们将遵循以下步骤 -
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更好地理解 -
#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}}输出结果12
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