[Template] minimum spanning tree [minimum spanning tree] [prim kruskal]

prim：

1. First establish a tree has only one node, this node can be original in any one node
2. The tree using an extended edge, this edge requires a vertex to another vertex in the tree, the tree is not, and for which the minimum requirement of the right side.
3. Repeat Step 2 until all vertices in the tree

Double Cheese seniors and see a template of the way to re-write the sensibility to understand the priority queue

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<cmath>
#include<stack>
#include<algorithm>
using namespace std;
#define ll long long
#define rg register
const int N=1000+5,M=1000000+5,inf=0x3f3f3f3f,P=19650827;
int n,m;
ll ans=0;
typedef pair<int,int>pii;                        //pair
priority_queue<pii,vector<pii>,greater<pii> >q; //小顶堆 
template <class t>void rd(t &x){
    x=0;int w=0;char ch=0;
    while(!isdigit(ch)) w|=ch=='-',ch=getchar();
    while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
    x=w?-x:x;
}

int tot=0,head[N];
struct edge{int v,nxt,w;}e[M<<1];
void add(int u,int v,int w){
    e[++tot]=(edge){v,head[u],w};head[u]=tot;
}

int say [N], screw [N];
void prim () {
    memset(vis,0,sizeof(vis));
    memset(dis,inf,sizeof(dis));
    q.push(make_pair(dis[1]=0,1));
    while(!q.empty()){
        int u=q.top().second;q.pop();
        if(vis[u]) continue;
        ans+=dis[u],vis[u]=1;
        for(int i=head[u];i;i=e[i].nxt){
            int v=e[i].v,w=e[i].w;
            if(!vis[v]&&dis[v]>w){
                dis[v]=w;
                q.push(make_pair(dis[v],v));
            }
        }
    }
}
int main () {
    rd (n), rd (m);
    int u,v,w;
    for(rg int i=1;i<=m;++i){
        rd (u), rd (v), rd (w);
        add(u,v,w),add(v,u,w);
    }
    prim();
    printf("%lld",ans);
    return 0;
}
 

P3366 [template] minimum spanning tree    plus a cnt to add a record number of points less than last n number indicates no connectivity

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<queue>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<stack>
 7 #include<algorithm>
 8 using namespace std;
 9 #define ll long long
10 #define rg register
11 const int N=5000+5,M=200000+5,inf=0x3f3f3f3f,P=19650827;
12 int n,m;
13 ll ans=0;
14 typedef pair<int,int>pii;                        //pair
15 priority_queue<pii,vector<pii>,greater<pii> >q; //小顶堆 
16 template <class t>void rd(t &x){
17     x=0;int w=0;char ch=0;
18     while(!isdigit(ch)) w|=ch=='-',ch=getchar();
19     while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
20     x=w?-x:x;
21 }
22 
23 int tot=0,head[N];
24 struct edge{int v,nxt,w;}e[M<<1];
25 void add(int u,int v,int w){
26     e[++tot]=(edge){v,head[u],w};head[u]=tot;
27 }
28 
29 int dis[N],vis[N],cnt=0;
30 bool prim(){
31     memset(vis,0,sizeof(vis));
32     memset(dis,inf,sizeof(dis));
33     q.push(make_pair(dis[1]=0,1));
34     while(!q.empty()){
35         int u=q.top().second;q.pop();
36         if(vis[u]) continue;
37         ans+=dis[u],vis[u]=1,++cnt;
38         for(int i=head[u];i;i=e[i].nxt){
39             int v=e[i].v,w=e[i].w;
40             if(!vis[v]&&dis[v]>w){
41                 dis[v]=w;
42                 q.push(make_pair(dis[v],v));
43             }
44         }
45     }
46     if(cnt!=n) return 0;
47     else return 1;
48 }
49 int main(){
50     rd(n),rd(m);
51     int u,v,w;
52     for(rg int i=1;i<=m;++i){
53         rd(u),rd(v),rd(w);
54         add(u,v,w),add(v,u,w);
55     }
56     if(!prim()) printf("orz");
57     else printf("%lld",ans);
58     return 0;
59 }
60  

3366 prim

 

kruskal

1. Figure T will follow the right side edge of small to large order, and the establishment of a no edge
   
2. There is no minimum elect a selected edge over the right side.
3. If this communication block edges in the two vertices where T is not the same, then it will be added to T in FIG.
4. FIG repeated until T 2 and 3 until the communication.

Since only need to maintain connectivity may not be required to establish the real FIG T, may be disjoint-set to maintain

Seniors code actually wrong! Shock!

Note BOOL operator <( const & Edge A) Finally, add const

Such written BOOL operator <( const & Edge A) const { return W <of Aw;}

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<cmath>
#include<stack>
#include<algorithm>
using namespace std;
#define ll long long
#define rg register
const int N=1000+5,M=1000000+5,inf=0x3f3f3f3f,P=19650827;
int n,m,cnt=0,f[N];
ll ans=0;

template <class t>void rd(t &x){
    x=0;int w=0;char ch=0;
    while(!isdigit(ch)) w|=ch=='-',ch=getchar();
    while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
    x=w?-x:x;
}

struct edge{
    int u,v,w;
    bool operator<(const edge&A)const{return w<A.w;}
}e[M];
int find(int x) {return f[x]==x?x:f[x]=find(f[x]);}

void kruskal(){
    for(rg int i=1;i<=n;++i) f[i]=i;
    for(rg int i=1,u,v;i<=m;++i){
        u = e [i] .u, v = e [i] .v;
        the if (the find (u)! = the find (v)) {
            f[f[u]]=f[v];
            ans+=e[i].w;
        }
    }
}

int main () {
    rd (n), rd (m);
    for(int i=1,u,v,w;i<=m;++i){
        rd (u), rd (v), rd (w);
        e[i]=(edge){u,v,w};
    }
    sort (e + 1 and + 1 + m);
    kruskal();
    printf("%lld",ans);
    return 0;
}