Problem Description

Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He needs your help, of course.
Farmer John ordered a high speed
connection for his farm and is going to share his connectivity with the other
farmers. To minimize cost, he wants to lay the minimum amount of optical fiber
to connect his farm to all the other farms.
Given a list of how much fiber
it takes to connect each pair of farms, you must find the minimum amount of
fiber needed to connect them all together. Each farm must connect to some other
farm such that a packet can flow from any one farm to any other farm.
The
distance between any two farms will not exceed 100,000.

 

Input

The input includes several cases. For each case, the
first line contains the number of farms, N (3 <= N <= 100). The following
lines contain the N x N conectivity matrix, where each element shows the
distance from on farm to another. Logically, they are N lines of N
space-separated integers. Physically, they are limited in length to 80
characters, so some lines continue onto others. Of course, the diagonal will be
0, since the distance from farm i to itself is not interesting for this
problem.

 

Output

For each case, output a single integer length that is
the sum of the minimum length of fiber required to connect the entire set of
farms.

 

Sample Input

 

Sample Output

 #include <iostream>

 #include <cstring>

 #include <string>

 #include <cstdio>

 #include <algorithm>

 using namespace std;

 int e[][];

 int d[];

 bool v[];

 int main()

 {

     int n, i, j;

     while (cin >> n)

     {

         for (i = ; i <= n; i++)

         {

             for (j = ; j <= n; j++)

             {

                 cin >> e[i][j];

             }

         }

         for (i = ; i <= n; i++)

         {

             d[i] = e[][i];

         }

         memset(v, , sizeof(v));

         v[] = ;

         int s = ;

         for (i = ; i <= n - ; i++)

         {

             int k = -;

             int mi = ;

             for (j = ; j <= n; j++)

             {

                 if (!v[j] && d[j] <= mi)

                 {

                     mi = d[j];

                     k = j;

                 }

             }

             if (k == -) break;

             v[k] = ;

             s = s + d[k];

             for (j = ; j <= n; j++)

             {

                 if (!v[j] && d[j] > e[k][j])//这一步和dijkstra不太同，不是d[j]>d[k]+e[k][j]

                 {

                     d[j] = e[k][j];

                 }

             }

         }

         cout << s << endl;

     }

     return ;

 }