Prim最小生成树算法

MST-PRIM(G, w, r)
   

// G: graph, w: weight,  r:root 
 1  for each u ∈ V [G]
   

 2       do key[u] ← ∞
   

 3          π[u] ← NIL
   

 4  key[r] ← 0
   

 5   Q ← V [G]
   

 6   while Q ≠ Ø
   

 7       do u ← EXTRACT-MIN(Q)
   

 8          for each v ∈ Adj[u]
   

 9              do if v ∈ Q and w(u, v) < key[v]
   

10                    then π[v] ← u
   

11                         key[v] ← w(u, v)

Prim's algorithm works as shown in the figure:

//  MST  set实现最小优先级队列
#include  < vector >
#include  < set >

int  MST_PRIM( const  vector < vector < pair < int ,  int >   >   >   &  G)
/*

• index start from 1 to n and
• G.size() -1 is the number of vertices in our graph
• G[i].size() is the number of vertices directly reachable from vertex with index i
• G[i][j].first is the index of j-th vertex reachable from vertex i
• G[i][j].second is the length of the edge heading from vertex i to vertex G[i][j].first 
  

         return the minimum cost of the spanning tree
*/
... {
    int n = G.size();
    vector<int> key (n, 987654321);
    vector<bool> belongQ(n, 1);
    key[1]=0;
    set<pair<int, int> > Q;
    for(int i=1; i<n; i++) 
        Q.insert(make_pair(key[i], i));
    while (!Q.empty()) ...{
        int u=Q.begin()->second;
        Q.erase(Q.begin());
        belongQ[u]=false;
        for(int i=0; i< G[u].size(); i++) ...{
            int v = G[u][i].first;
            if(belongQ[v] && G[u][i].second < key[v]) ...{
                Q.erase(Q.find(make_pair(key[v], v)));
                key[v]=G[u][i].second;
                Q.insert(make_pair(key[v], v));
            }
        }
    } 
    int res=0;
    for(int i=1; i < n; i++) 
        res+=key[i];
    return res;
}

 
题目：http://acm.pku.cn/JudgeOnline/problem?id=1251 jungle roads   
            http://acm.pku.cn/JudgeOnline/problem?id=1287 Networking