The minimum spanning tree problem ----- P3366 template [template] minimum spanning tree

Title Description

If that, given an undirected graph, the minimum spanning tree is obtained, if this figure is not connected, the output orz

Input Format

The first line contains two integers N, M, N represents the total FIG nodes and the M undirected edges. (N <= 5000, M <= 200000)

Next M lines contains three integers Xi, Yi, Zi, Zi expressed a length of edges connected to node No Xi, Yi

Output Format

Comprising a number of output, i.e., the minimum spanning tree and the length of each side; if the output is not connected orz FIG.

Sample input and output

Input # 1

Output # 1

Description / Tips

Constraints of time: 1000ms, 128M

Data Scale:

For 20% of the data: N <= 5, M <= 20

For 40% of the data: N <= 50, M <= 2500

For 70% of the data: N <= 500, M <= 10000

To 100% of the data: N <= 5000, M <= 200000

Sample explained:

Therefore, the minimum spanning tree of total edge weight of 2 + 2 + 3 = 7

Ideas: an input side (left points a, a right end points b, the length w) all fit inside the vector, and then by the length from small to large, after traversing the vector, using disjoint-set ab determines whether the endpoint is connected, has been linked to skipped, otherwise to use (a, b, w) connect.
Finally, connected to the N-1 when the edges, is already FIG communication, the edge is small and is selected to the election, and so is the minimum length.
If the edge is not selected edges to N-1, then no communication is definitely FIG.

AC codes and templates

#include <bits / STDC H ++.>
 the using  namespace STD;
 const  int NN = 1E4 + 10 ;
 int N, M;
 int FA [NN];
 struct Node {
     int A, B, W; 
    Friend BOOL  operator <(Node N1, Node N2) { // collation node structure, not want to define cmp, after a write function 
        return n1.w < n2.w; 
    } 
}; 
Vector <node> VE; // storage container side 

int Find ( int X ) { // Find ancestors 
    IF (the X-! = FA [the X-]) 
        FA [the X-] = find(fa[x]);
    return fa[x]; 
}

void join(int x,int y){ //合并 
    int fx = find(x),fy = find(y);
    if(fx!=fy)
        fa[fx] = fy;
}


int main(){
    cin>>N>>M;
    for(int i = 1;i<=N;i++) fa[i] = i;
    int a,b,w;
    while(M--){
        scanf("%d%d%d",&a,&b,&w);
        ve.push_back ({A, B, W}); 
    } 
    Sort (ve.begin (), ve.end ()); // Sort 
    Long  Long SUM = 0 ; // All patterned edges total length 
    int Edges = 0 ; // selected number of sides 
    for (Auto V: VE) {
         IF {(Find (VA) = Find (VB)!) // if the point is not a point b Unicom, chose this edge merge 
            join (va, vb ); 
            SUM + = VW; 
            Edges ++ ; 
        } 
    } 
    IF (Edges> = N- . 1 ) SUM COUT << << endl;
     the else COUT <<"orz"<<endl;
    
    return 0; 
}