题目:
刚开始一看以为这是货郎担(哈密顿环)问题,后来发现并不是这样。因为货郎担问题中所有的顶点只能走一次,美团的这个题目是可以重复访问的,只要最终的路程最短即可。并且货郎担问题需要重新回到起点,但这个图的遍历并不要求。
C++实现的代码:
#include <iostream>
#include <vector>
#include<algorithm>
using namespace std;
#define max_num 100000
int cost(int i, int j, int n, int **arr)
{
int w_sum = 0;
while (i < n - 1 && j < n)
{
if (i!=j && arr[i][j]!=max_num)
{
w_sum += 1;
i = j;
//visited[j] = false;
j++;
}
else
{
j++;
}
}
return w_sum;
}
int main()
{
int N;
int link_num=0;
int *costed;
cin >> N;
//bool visited[max_num];
int **arr = new int*[N];
for (int i = 0; i < N; i++)
{
arr[i] = new int[N];
for (int j = 0; j < N; j++)
{
if (i == j)
{
arr[i][i] = 0;
}
else
{
arr[i][j] = max_num;
}
}
}
for (int i = 0; i < N-1; i++)
{
int p, q;
cin >> p >> q;
arr[p - 1][q - 1] = arr[q - 1][p - 1] = 1;
}
for (int i = 0; i < N; i++)
{
if (arr[0][i]==1)
{
link_num += 1;
}
}
costed = new int[N];
for (int i = 0; i < N; i++)
{
if (arr[0][i] == 1)
{
costed[i] = cost(0, i, N, arr);
}
else
{
costed[i] = 0;
}
}
sort(costed,costed+N);
int short_sum=0;
for (int i = 0; i < N; i++)
{
if (i<N-1)
{
short_sum += costed[i] * 2;
}
else
{
short_sum += costed[i];
}
}
cout << short_sum;
return 0;
}