LuoGu P1547 Out of Hay's explanations

(* ^ - ^ *) hope Gangster pointing problem solution of konjac (* ^ __ ^ *)

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I personally think that this question is the template title Kruskal algorithm is quite simple .

The Kruskal (Kruskal) : using an ingenious disjoint-set to the minimum spanning tree algorithm.

Implementation process:

Process is as follows:

1. Initialization, initialize the disjoint-set, each point of the parent node for itself.
2. Counter to zero.
3. All right edge of the value of small to large fast row.
4. Specific Kruskal's algorithm:

 1 for(int i=1;i<=m;i++)
 2 {
 3     int f1=Find(e[i].x),f2=Find(e[i].y);
 4     if(f1!=f2)
 5     {
 6         ans=max(ans,e[i].dis);
 7         fa[f1]=f2;
 8         if(k==n-1) break;
 9         k++;
10     }
11 }

 

Full code below (comment Diudiu a)

code

#include <bits / STDC ++ H.>
 the using  namespace STD;
 int n-, m, ANS, K = . 1 , FA [ 100005 ];
 struct Node // represented by edge structure 
{
     int X, Y, DIS; 
} E [ 100005 ];
 BOOL CMP (Node x, Node Y) 
{ 
    return x.dis < y.dis; 
} 
int the Find ( int x) // find the ancestors of x 
{
     IF FA [x] = (FA [x] = x!) the Find (FA [X]);
     return FA [X]; 
} 
void of Union ( int X, int y)//将x,y并为一个集合
{
    if(Find(x)!=Find(y))
        fa[Find(x)]=Find(y);
}
int main()
{
    cin>>n>>m;
    for(int i=1;i<=n;i++)
        fa[i]=i;
    for(int i=1;i<=m;i++)
        cin>>e[i].x>>e[i].y>>e[i].dis;
    sort(e+1,e+m+1,cmp);//快排
    for(int I = . 1 ; I <= m; I ++ ) 
    { 
        int F1 = the Find (E [I] .x), F2 = the Find (E [I] .y);
         IF ! (F1 = F2) // if different fathers 
        { 
            ANS = max (ANS, E [I] .dis); // update answer 
            FA [F1] = F2;
             IF (K == N- . 1 ) BREAK ; // least sides connected points -1 
            K ++; // counter ++ 
        } 
    } 
    COUT << ANS << endl; // output 
    return  0 ; 
}

 

 

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