最小生成树 —— 模板

时间复杂度对比：
Prim：$O(n^2)$             适用于稀疏图
Prim + 优先队列（堆优化）：$O((n+m)\ast logm)$
Kruskal：$O(m\ast logm+m\alpha (n))$，$\alpha (n)$为常数        适用于稠密图

$PS$：文末附三者算法效率分析$......$

---

Prim + 邻接链表：

#include<bits/stdc++.h>
#define rint register int
#define deb(x) cerr<<#x<<" = "<<(x)<<'\n';
using namespace std;
typedef long long ll;
typedef pair <int,int> pii;
const ll mod = 1e9 + 7;
const int maxn = 2e5 + 10;
int n, m, cnt, ans, dis[maxn];
int head[maxn], vis[maxn]; 
struct EDGE{
	int next, to, w;  
} e[maxn<<1]; 

inline int read(){
    int x=0;
    char ch=getchar();
    while(ch>'9' || ch<'0') ch=getchar();
    while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
    return x;
}

void add(int u, int v, int w) {  
	e[++cnt].next = head[u];
	e[cnt].w = w;
	e[cnt].to = v;
	head[u] = cnt;    
}

void prim(int st){
	memset(vis,0,sizeof(vis));
	for(int i=1; i<=n; i++) dis[i] = INT_MAX;
	for(int i=head[st]; i; i=e[i].next)
		dis[e[i].to] = min(dis[e[i].to], e[i].w);
	dis[st] = 0, vis[st] = 1;
	for(int i=1; i<n; i++){
		int tmp = INT_MAX, k = 0;
		for(int j=1; j<=n; j++)
			if(!vis[j] && tmp>dis[j]){
				k = j;
				tmp = dis[j];
			}
		vis[k] = 1;
		ans += tmp;
		for(int j=head[k]; j; j=e[j].next)
			if(!vis[e[j].to] && dis[e[j].to]>e[j].w)
				dis[e[j].to] = e[j].w;
	}
}

int main() {
	n = read(), m = read();
	memset(head,0,sizeof(head));
	ans = cnt = 0;
	for(int i=0; i<m; i++){
		int u, v, w;
		u = read(), v = read(), w = read();
		add(u, v, w);
		add(v, u, w);
	}
	prim(1);
	printf("%d\n", ans);
}

---

Prim + 邻接链表 + 堆优化（优先队列）：

#include<bits/stdc++.h>
#define rint register int
#define deb(x) cerr<<#x<<" = "<<(x)<<'\n';
using namespace std;
typedef long long ll;
typedef pair <int,int> pii;
const ll mod = 1e9 + 7;
const int maxn = 2e5 + 10;
int n, m, cnt, ans, vis[maxn];
int head[maxn], dis[maxn]; 
priority_queue <pii, vector<pii>, greater<pii> > q;
struct EDGE{
	int next, to, w;  
} e[maxn<<1]; 

inline int read(){
    int x=0;
    char ch=getchar();
    while(ch>'9' || ch<'0') ch=getchar();
    while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
    return x;
}

void add(int u, int v, int w) {  
	e[++cnt].next = head[u];
	e[cnt].w = w;
	e[cnt].to = v;
	head[u] = cnt;    
}

void prim(int st){
	memset(vis,0,sizeof(vis));
	for(int i=1; i<=n; i++) dis[i] = INT_MAX;
	q.push(make_pair(0, st));
	dis[st] = 0;
	int num = 0;
	while(!q.empty() && num<n){
		int d = q.top().first;
		int k = q.top().second;
		q.pop();
		if(vis[k]) continue;
		num++;
		ans += d;
		vis[k] = 1;
		for(int i=head[k]; i; i=e[i].next)
			if(e[i].w<dis[e[i].to]){
				dis[e[i].to] = e[i].w;
				q.push(make_pair(e[i].w, e[i].to));
			}
	}
}

int main() {
	n = read(), m = read();
	memset(head,0,sizeof(head));
	ans = cnt = 0;
	for(int i=0; i<m; i++){
		int u, v, w;
		u = read(), v = read(), w = read();
		add(u, v, w);
		add(v, u, w);
	}
	prim(1);
	printf("%d\n", ans);
}

---

Kruskal：

#include<bits/stdc++.h>
#define rint register int
#define deb(x) cerr<<#x<<" = "<<(x)<<'\n';
using namespace std;
typedef long long ll;
typedef pair <int,int> pii;
const ll mod = 1e9 + 7;
const int maxn = 2e5 + 10;
int n, m, cnt, ans, f[maxn]; 
struct EDGE{
	int u, v, w;
	bool operator <(const EDGE &A) const{
		return w<A.w;
	}
} e[maxn];

inline int read(){
    int x=0;
    char ch=getchar();
    while(ch>'9' || ch<'0') ch=getchar();
    while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
    return x;
}

int find(int x){
	return x == f[x] ? x : f[x] = find(f[x]);
}

int main() {
	int u, v, w;
	n = read(), m = read();
	for(int i=1; i<=n; i++) f[i] = i;
	for(int i=0; i<m; i++){
		e[i].u = read();
		e[i].v = read();
		e[i].w = read();
	}
	sort(e, e+m);
	for(int i=0; i<m; i++){
		u = find(e[i].u);
		v = find(e[i].v);
		if(u!=v){
			f[u] = v;
			ans += e[i].w;
			cnt++;
		}
		if(cnt==n-1) break;
	}
	printf("%d\n", ans);
}

---

附上Prim，Prim+heap，Kruskal算法效率分析 （用堆优化就完事了呗）: