N - Is There A Second Way Left? (minimum spanning tree, Kruskal)

Nasa, being the most talented programmer of his time, can’t think things to be so simple. Recently all
his neighbors have decided to connect themselves over a network (actually all of them want to share
a broadband internet connection ). But he wants to minimize the total cost of cable required as he
is a bit fastidious about the expenditure of the project. For some unknown reasons, he also wants a
second way left. I mean, he wants to know the second best cost (if there is any which may be same as
the best cost) for the project. I am sure, he is capable of solving the problem. But he is very busy with
his private affairs(?) and he will remain so. So, it is your turn to prove yourself a good programmer.
Take the challenge (if you are brave enough)…
Input
Input starts with an integer t ≤ 1000 which denotes the number of test cases to handle. Then follows
t datasets where every dataset starts with a pair of integers v (1 ≤ v ≤ 100) and e (0 ≤ e ≤ 200). v
denotes the number of neighbors and e denotes the number of allowed direct connections among them.
The following e lines contain the description of the allowed direct connections where each line is of the
form ‘start end cost’, where start and end are the two ends of the connection and cost is the cost for
the connection. All connections are bi-directional and there may be multiple connections between two
ends.
Output
There may be three cases in the output

1. No way to complete the task,
2. There is only one way to complete the task,
3. There are more than one way.
   Output ‘No way’ for the first case, ‘No second way’ for the second case and an integer c for the
   third case where c is the second best cost. Output for a case should start in a new line.
   Sample Input
   4
   5 4
   1 2 5
   3 2 5
   4 2 5
   5 4 5
   5 3
   1 2 5
   3 2 5
   5 4 5
   5 5
   1 2 5
   3 2 5
   4 2 5
   5 4 5
   4 5 6
   1 0
   Sample Output
   Case #1 : No second way
   Case #2 : No way
   Case #3 : 21
   Case #4 : No second way
   The question ：
   Output three results ：
   No shortest circuit , Output No way
   There is only one output No second way
   There are multiple paths to output the second shortest circuit ,
   And the o Basically the same , hold o The code of the question has been changed , Just accept 了

#include <cstdio>
#include <cmath>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N=888;
int MAX=0x3f3f3f3f;
int n,m;
int dist[N],map[N][N];
int biao[N];
int sum[N];
int x[N],y[N];
int mm[N];
int pre[N];
struct node{
    
	int u,v,w;
}a[N*(N-1)/2];
bool cmp(node a,node b)
{
    
	return a.w<b.w;
}
int find(int x)
{
    
	if(pre[x]==x)
	return x;
	else
	return pre[x]=find(pre[x]);
}
int judge(int u,int v)
{
    
	int fx=find(u);
	int fy=find(v);
	if(fx!=fy)
	{
    
		pre[fy]=fx;
		return 1;
	}
	return 0;
}
void unit()
{
    
	for(int i=1;i<=n;i++)
	{
    
		pre[i]=i;
	}
}
int main()
{
    
   int t;
   cin>>t;
   int sum=0;;
   while(t--)
   {
    
   	sum++;
   	cin>>n>>m;
   	if(n==1&&m==0)// A special case , Discuss alone  
   	cout<<"Case #"<<sum<<" : No second way"<<endl;
   	else
   	{
    
   			unit();
   	memset(biao,0,sizeof(biao));
   	for(int i=0;i<m;i++)
   	{
    
   	  cin>>a[i].u>>a[i].v>>a[i].w;
	}
	sort(a,a+m,cmp);
	int ans=0,num=0;
	int flag=0;// Mark , Used to determine whether a complete path can be formed  
	for(int i=0;i<m;i++)
	{
    
		if(judge(a[i].u,a[i].v))
		{
    
			ans+=a[i].w;
			num++;
			biao[i]=1;
			if(num==n-1)
			{
    
				flag=1;// There is a minimum spanning tree ,
				break;
			}
		}
	}
	int minn=ans;
	cout<<"Case #"<<sum<<" : ";
	if(flag==0)
	cout<<"No way"<<endl;
	else
	{
    
		flag=0;// Restart marking 
		int minnn=MAX;
	for(int i=0;i<m;i++)
	{
    
		ans=0,num=0;
		if(biao[i]==1)
		{
    
			unit();
			for(int j=0;j<m;j++)
			{
    
				if(i==j)
				continue;
		    if(judge(a[j].u,a[j].v))
	    	{
    
			ans+=a[j].w;
			num++;
			if(num==n-1)
			{
    
				flag=1;// Is there any tree smaller than the minimum spanning tree 
				minnn=min(minnn,ans);// Find a small spanning tree 
					break;
			}
	    	}
			}
		}
	}
	if(flag==0)
	cout<<"No second way"<<endl;
	else
	   cout<<minnn<<endl;
	}
	   }
   }
    return 0;
}