Problem:
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Minimum Spanning Tree

For a given weighted graph $G = (V, E)$, find the minimum spanning tree (MST) of $G$ and print total weight of edges belong to the MST.

Input

In the first line, an integer $n$ denoting the number of vertices in $G$ is given. In the following $n$ lines, a $n \times n$ adjacency matrix $A$ which represents $G$ is given.
$a_{ij}$ denotes the weight of edge connecting vertex $i$ and vertex $j$. If there is no edge between $i$ and $j$, $a_{ij}$ is given by -1.

Output

Print the total weight of the minimum spanning tree of $G$.

Constraints

$1 \leq n \leq 100$

$0 \leq a_{ij} \leq 2,000$ (if $a_{ij} \neq -1$)

$a_{ij} = a_{ji}$

$G$ is a connected graph

Sample Input 1

5
-1 2 3 1 -1
2 -1 -1 4 -1
3 -1 -1 1 1
1 4 1 -1 3
-1 -1 1 3 -1

Sample Output 1

5

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in C (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#define MAX 100
#define number 2000

int A[MAX][MAX];
int nord[MAX];
int start = 0;

int searchmin(int n, int *C) {
   int i, j;
   for (i = start; i < n; i++) {
	  for (j = i + 1; j < n; j++) {
		 if (A[i][j] == *C && (nord[i] == 0 && nord[j] == 1 || nord[j] == 0 && nord[i] == 1)) {
			nord[i] = 1;
			nord[j] = 1;
			return A[i][j];
		 }
	  }
   }
   for (i = 0; i < start; i++) {
	  for (j = i + 1; j < n; j++) {
		 if (A[i][j] == *C && (nord[i] == 0 && nord[j] == 1 || nord[j] == 0 && nord[i] == 1)) {
			nord[i] = 1;
			nord[j] = 1;
			return A[i][j];
		 }
	  }
   }
   start = 0;
   *C = *C + 1;
   return searchmin(n, C);
}

int main() {
   int n, ans = 0;
   int check = 0, cost;
   int i, j;

   scanf("%d", &n); 

   for (i = 0; i < n; ++i) {
	  nord[i] = 0;
	  for (j = 0; j < n; ++j) {
		 scanf("%d", &A[i][j]);
	  }
   }

   nord[0] = 1;
   while (check == 0) {
	  ans += searchmin(n, &cost);
	  check = 1;
	  for (i = 0; i < n; ++i) {
		 check = check * nord[i];
		 if (nord[i] == 0)  break;
	  }
   }
   printf("%d\n", ans);
   return 0;
}