最小生成树(prim算法)

1.朴素prim算法(重在理解prim算法思想)

#include 
using namespace std;
using ll = long long;
const int N = 300 + 9;
const ll inf = 4e18;

ll a[N][N], d[N], n, m;

bitset intree;

void solve()
{
   cin >> n >> m ;

   memset(a,0x3f,sizeof a);
   memset(d,0x3f,sizeof d);

   for(int i = 1;i <= n;i++)a[i][i] = 0;

   for(int i = 1;i <= m;i++)
   {
       ll u , v , w;cin >> u >> v >> w;
       // 这个是无向图啊
       // 所以这里两边都要更新一下子
       a[u][v] = min(a[u][v] , w);
       a[v][u] = min(a[v][u] , w);
   }

   /*
   d[1] = 0; // 1 放入了intree集合(intree 和 outtree 都不是实际创建的容器集合)
   intree[1] = true;
   */

   ll ans = 0;

   for(int i = 1;i <= n;i++)
   {
       // 为了找出与 intree最新点 距离最近的下一点
       int u = 1;
       for(int j = 1;j <= n;j++)
           if(intree[u] || (!intree[j] && d[j] < d[u]))u = j; 

       if(u ^ 1)ans += d[u];
       intree[u] = true;
       d[u] = 0;

       for(int j = 1;j <= n;j++)
       {
           if(intree[j])continue;

           d[j] = min(a[u][j],d[j]);
       }
   }

   cout << ans << '\n';

}

int main()
{
   ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);

   int _ = 1;
   while(_--)
       solve();

   return 0;
}

2.用优先队列优化prim算法

#include
using namespace std;
using ll=long long;
const int N=1e5+9;
ll d[N];
bitsetintree;
struct Edge
{
  ll x,w;
  bool operator <(const Edge &e)const
  {
    return w==e.w?xe.w;
  }
};
vectorg[N];
int main()
{
    memset(d,0x3f,sizeof d);
    ll n,m;cin>>n>>m;
    for(int i=1;i<=m;i++)
    {
        ll u,v,w;cin>>u>>v>>w;
        g[u].push_back({v,w});
        g[v].push_back({u,w});
    }
    priority_queuepq;
    ll ans=0;
    pq.push({1,0});
    d[1]=0;
    while(pq.size())
    {
        auto[x,w]=pq.top();pq.pop();
        if(intree[x])continue;
        intree[x]=true;
        ans+=w;
        d[x]=0;
        for(auto &[y,w]:g[x])
        {
          if(!intree[y]&&w

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