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 NOTSIDE -1
#define WHITE 0
#define GRAY 1
#define BLACK 2
static const int INFTY = (1<<21);

int A[MAX][MAX],n,v,i;

int prim(){
  int u,mincost,sum=0;
  int d[MAX],p[MAX],color[MAX];
 
  for(i=0;i<n;i++){
    d[i]=INFTY;
    p[i]=NOTSIDE;
    color[i]=WHITE;
  }
 
  d[0]=WHITE;
  p[0]=NOTSIDE;
 
  while(1){
    mincost=INFTY;
    u=NOTSIDE;
    for(i=0;i<n;i++){
      if(mincost>d[i]&&color[i]!=BLACK){
        u=i;
        mincost=d[i];
      }
    }
    if(u==INFTY)  break;
    color[u]=BLACK;
    for(v=0;v<n;v++){
      if(color[v]!=BLACK&&A[u][v]!=INFTY){
        if(d[v]>A[u][v]){
          d[v]=A[u][v];
          p[v]=u;
          color[v]=GRAY;
        }
      }
    }
  }
  for(i=0;i<n;i++){
    if(p[i]!=NOTSIDE) sum+=A[i][p[i]];
  }
  return sum;
}

int main(){
  int e,j;
  scanf("%d",&n);
  for(i=0;i<n;i++){
    for(j=0;j<n;j++){
      scanf("%d",&A[i][j]);
      A[i][j]=(e==NOTSIDE)?INFTY:e;
    }
  }
  printf("%d\n",prim());
  return 0;
}