hdu1233最小生成树prim算法模版题

#include<iostream>
#include<vector>
#include<queue>
#include<cstdio>
using namespace std;

struct edge
{
    int from,to,cost;
    friend bool operator <(edge e1,edge e2)
    {
        return e1.cost>e2.cost;
    }
};
const int MAXN=105;
vector<edge>v[MAXN];
bool selected[MAXN];
int prim(int p)
{
    priority_queue<edge>q;
    int sum=0;
    selected[p]=1;
    int i=0;
    while(!q.empty())
    {
        q.pop();
    }
    for(i=0;i<=int(v[p].size())-1;i++)
    {
        int to=v[p][i].to;
        if(!selected[to])
        {
            q.push(v[p][i]);
        }
    }
    while(!q.empty())
    {
        int from;
        int i=0;
        int to;
        edge e=q.top();
        to=e.to;
        from=e.from;
        q.pop();
        if(!selected[to])
        {
            sum=sum+e.cost;
            selected[e.to]=1;
            for(i=0;i<=int(v[to].size())-1;i++)
            {
                int too=v[to][i].to;
                if(!selected[too])
                {
                    q.push(v[to][i]);
                }
            }
        }
    }
    return sum;
}
void init(int n)
{
    int i=0;
    for(i=0;i<=n;i++)
    {
        v[i].clear();
    }
    memset(selected,0,sizeof(selected));
}
int main()
{
    int n;
    while((scanf("%d",&n))&&(n!=0))
    {
        int from,to,cost;
        init(n);
        n=(n*(n-1))/2;
        while(n--)
        {
            scanf("%d%d%d",&from,&to,&cost);
            edge e={from,to,cost};
            v[e.from].push_back(e);
            swap(e.from,e.to);
            v[e.from].push_back(e);
        }
        cout<<prim(1)<<endl;
        
    }
    return  0;
}

  

原文地址:https://www.cnblogs.com/cj695/p/2609493.html