acm algorithm practice Dec. 27 MST

what makes me amused joyful is I use emacs coding this algorithm..:)

although i am not proficient in it at all.

besides, today I installed aqumacs in place of the built-in emacs of Mac.   thereby（thus）  reviewed  the bash command and the profile .bash_profile

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poj 1258   Agri-Net

Input

The input includes several cases. For each case, the first line contains the number of farms, N (3 <= N <= 100). The following lines contain the N x N conectivity matrix, where each element shows the distance from on farm to another. Logically, they are N lines of N space-separated integers. Physically, they are limited in length to 80 characters, so some lines continue onto others. Of course, the diagonal will be 0, since the distance from farm i to itself is not interesting for this problem.

Output

For each case, output a single integer length that is the sum of the minimum length of fiber required to connect the entire set of farms.

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it's a basic MST problem using Kruskac's algorithm.   (since Prime algorithm is kinda like djstrak algorithm which is practiced yesterday :)).   

both the data structure and the algorithm are quite simple.    

the key of K algorithm is Union-find  &  union

the code is:   

  edges.sort(cmp);   

  for(list<edge>::iterator it = edges.begin(); it != edges.end(); it++){
    //if find set
    int x = (*it).from;
    int y = (*it).to;
    while(vertices[x].parent != x){
      x = vertices[x].parent; // climb up
      
    }
    while(vertices[y].parent != y){
      y = vertices[y].parent;  //climb up
    }
    
    if(x == y){//reject 
      continue;
    }else{
      //put the edge into A 
      A += (*it).weight;
      //flip, now the x, y is the ROOT !!!!!
      
      if(vertices[x].size <= vertices[y].size){
    vertices[x].parent = y;   // x's parent is set to y 
    vertices[y].size = vertices[x].size + vertices[y].size;
      }else{
    vertices[y].parent = x;   // y's parent is set to x 
    vertices[x].size = vertices[x].size + vertices[y].size;
      }    
    }
    //else do nothing 
  }

 

the caveat doesn't belong to the algorithm issue... at first,  I can not pass poi until I change the input 

while (cin >> N) {}

 

.  the instruction itself doesn't mention this input format :( and thank God I didn't change my original implementation of algorithm, or it would be quite a waste of time ....

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