UVA 216 - Getting in Line

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题目大意：把所有点用一条线联通，求出所有这些线中长度最短的一条

解题思路：求出这些点的全排列，并计算它的长度，记录最短长度的那排列

#include <cstdio>#include <algorithm>#include <cmath>using namespace std;int main() {int num, arr[10][2], brr[10], order[10], count = 0;while (scanf("%d", &num), num) {double min = 1000000;for (int i = 0; i < num; i++)scanf("%d%d", &arr[i][0], &arr[i][1]), brr[i] = i;do {double sum = 0;for (int i = 0; i < num - 1; i++) {int x = arr[brr[i]][0] - arr[brr[i + 1]][0];int y = arr[brr[i]][1] - arr[brr[i + 1]][1];sum += sqrt(x * x + y * y);}if (sum < min) {min = sum;for (int i = 0; i < num; i++)order[i] = brr[i];}} while (next_permutation(brr, brr + num)); //寻找出所有点的全排列中距离之和最小的序列并记录下来printf("**********************************************************\nNetwork #%d\n", ++count);for (int i = 0; i < num - 1; i++) {int x = arr[order[i]][0] - arr[order[i + 1]][0];int y = arr[order[i]][1] - arr[order[i + 1]][1];printf("Cable requirement to connect (%d,%d) to (%d,%d) is %.2f feet.\n", arr[order[i]][0], arr[order[i]][1], arr[order[i + 1]][0], arr[order[i + 1]][1], sqrt(x * x + y * y) + 16.0);}printf("Number of feet of cable required is %.2f.\n", min + 16 * (num - 1));}return 0;}

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