【模板】 最小生成树

 prim：
给出一个无向图，求出最小生成树，如果该图不连通，则输出impossible。

#include<bits/stdc++.h>
using namespace std;
const int maxm = 200005;
const int inf = 0x3f3f3f3f;
struct edge{
 int v, w, next;
}e[maxm << 1];
int cnt, n, m, sum, t, p, now = 1;
int dis[5005], head[maxm], vis[5005];
int a, b, c;

void add(int uu, int vv, int ww)
{
 e[++cnt].v = vv;
 e[cnt].w = ww;
 e[cnt].next = head[uu];
 head[uu] = cnt;
}
int main()
{
 cin >> n >> m;
 for(int i = 1; i <= m; i++)
 {
  cin >> a >> b >> c;
  add(a, b, c); add(b, a, c); // 无向图，所以两个点之间连上 
 }
 for(int i = 1; i <= n; i++)
  dis[i] = inf;
 for(int i = head[1]; i; i = e[i].next) // 判断重边 
  dis[e[i].v] = min(e[i].w, dis[e[i].v]);
 while(++t < n)
 {
  vis[now] = 1;
  int minn = inf;
  now = -1;
  for(int i = 1; i <= n; i++)
  {
   if(!vis[i] && minn > dis[i])
   {
    minn = dis[i];
    now = i;
   }
  }
  if(now == -1) // 若有一个点更新完后dis[]数组仍为 inf，那么不能构成最小生成树 
  {
   cout << "impossible";
   return 0;
  }
  sum += minn;
  for(int i = head[now]; i; i = e[i].next) //更新dis[]数组 
  {
   if(!vis[e[i].v] && e[i].w < dis[e[i].v])
    dis[e[i].v] = e[i].w;
  }
 }
 cout << sum;
 return 0;
}

 

给出一个无向图，求出最小生成树。不用判断是否连通

#include<bits/stdc++.h>
using namespace std;
const int maxm = 200005;
const int inf = 0x3f3f3f3f;
struct edge{
    int v, w, next;
}e[maxm << 1];
int cnt, n, m, sum, t, p, now = 1;
int dis[5005], head[maxm], vis[5005];
int a, b, c;

void add(int uu, int vv, int ww)
{
    e[++cnt].v = vv;
    e[cnt].w = ww;
    e[cnt].next = head[uu];
    head[uu] = cnt;
}
int main()
{
    cin >> n >> m;
    for(int i = 1; i <= m; i++)
    {
        cin >> a >> b >> c;
        add(a, b, c); add(b, a, c);
    }
    for(int i = 1; i <= n; i++)
        dis[i] = inf;
    for(int i = head[1]; i; i = e[i].next)
        dis[e[i].v] = min(e[i].w, dis[e[i].v]);
    while(++t < n)
    {
        vis[now] = 1;
        int minn = inf;
        for(int i = 1; i <= n; i++)
        {
            if(!vis[i] && minn > dis[i])
            {
                minn = dis[i];
                now = i;
            }
        }
        if(dis[now] == inf) 
        {
            cout << "orz";
            return 0;
        }
        sum += minn;
        for(int i = head[now]; i; i = e[i].next)
        {
            if(!vis[e[i].v] && e[i].w < dis[e[i].v])
                dis[e[i].v] = e[i].w;
        }
    }
    cout << sum;
    return 0;
}

两者差别在于每次循环边时对now的处理