K - or unblocked project (minimum spanning tree)

Investigation of rural traffic in a province , The resulting table shows the distance between any two villages . provincial government “ Unimpeded works ” Our goal is to achieve road traffic between any two villages in the province （ But it doesn't have to be directly connected by road , As long as it can be reached indirectly by road ）, And the total length of the road is required to be the minimum . Please calculate the minimum total length of the highway .
Input
The test input contains several test cases . The... Of each test case 1 Row gives the number of villages N ( < 100 ); And then N(N-1)/2 The row corresponds to the distance between villages , Each line gives a pair of positive integers , They are the numbers of the two villages , And the distance between the two villages . For the sake of simplicity , Village from 1 To N Number .
When N by 0 when , End of input , The use case is not processed .
Output
For each test case , stay 1 The minimum total length of the highway output in the line .
Sample Input
3
1 2 1
1 3 2
2 3 4
4
1 2 1
1 3 4
1 4 1
2 3 3
2 4 2
3 4 5
0
Sample Output
3
5

Huge input, scanf is recommended.
Hint
Hint
// The template questions

#include <cstdio>
#include <cmath>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N=200,MAX=0x3f3f3f3f;
int n;
int from,to,w;
int dist[N],map[N][N];
bool biao[N];// Mark 
int prim()
{
    
	memset(dist,0x3f,sizeof(dist));
	memset(biao,0,sizeof(biao));
	int t=1,sum=0;
	dist[1]=0,biao[1]=1;
	for(int i=0;i<n-1;i++)
	{
    
		for(int j=1;j<=n;j++)
		dist[j]=min(dist[j],map[t][j]);
		t=-1;
		for(int j=1;j<=n;j++)
		{
    
			if(!biao[j]&&(t==-1||dist[t]>dist[j]))
			t=j;
		}
		sum+=dist[t];
		biao[t]=1;
	}
	return sum;
}
int main()
{
    
  while(~scanf("%d",&n))
  {
    
  	if(n==0)
  	break;
  	int m=n*(n-1)/2;
  memset(map,0x3f,sizeof(map));
  while(m--)
  {
    
  // cin>>from>>to>>w;
  	scanf("%d%d%d",&from,&to,&w);
  	map[from][to]=map[to][from]=min(map[from][to],w);
  }
  int ans=prim();
  cout<<ans<<endl;
  }
  
    return 0;
}

Detailed description and explanation of minimum spanning tree