UVa 216 Getting in Line

旅行商TSP(traveling salesmen problem)是一个NP难问题，至今没有有效的解决方案。本问题的规模为8！= 40320种可能，故使用枚举解决，这里枚举通过使用STL的next_permutation()函数来实现。AC代码如下：

/* coder: ACboy date: 2010-3-19 result: AC description: UVa 216 Getting in Line */ #include <iostream> #include <cmath> using namespace std; struct node { double x; double y; }; struct Dis { node a; node b; double dis; }; // 用于保存结果 Dis ans[10]; // 用于保存中间结果 Dis tempAns[10]; // 用于保存输入的节点 node in[10]; int n; // 求两点之间的距离 double getDis(node & a, node & b) { return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y)) + 16; } int main() { #ifndef ONLINE_JUDGE freopen("216.txt", "r", stdin); #endif int count = 0; while (cin >> n) { if (n == 0) break; int i; int data[10]; for (i = 0; i < n; i++) { cin >> in[i].x >> in[i].y; data[i] = i; } double min = 1 << 30; int c; // 遍历每一个可能。当n = 8时一共有40320种方案 do { double sum = 0; c = 0; for (i = 0; i < n - 1; ++i) { double tempDis = getDis(in[data[i]], in[data[i + 1]]); tempAns[c].a = in[data[i]]; tempAns[c].b = in[data[i + 1]]; tempAns[c].dis = tempDis; c++; sum += tempDis; } if (sum < min) { min = sum; for (i = 0; i < c; i++) { ans[i].a = tempAns[i].a; ans[i].b = tempAns[i].b; ans[i].dis = tempAns[i].dis; } } } while (next_permutation(data, data + n)); cout << "**********************************************************" << endl; cout << "Network #" << ++count << endl; for (i = 0; i < c; i++) { printf("Cable requirement to connect (%.0lf,%.0lf) to (%.0lf,%.0lf) is %.2lf feet./n", ans[i].a.x, ans[i].a.y, ans[i].b.x, ans[i].b.y, ans[i].dis); } printf("Number of feet of cable required is %.2lf./n", min); } return 0; }