模板 最小生成树

Posted 2022-12-10 lovezxy520

 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);
    
    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;
        now = -1;
        for(int i = 1; i <= n; i++)
        
            if(!vis[i] && minn > dis[i])
            
                minn = dis[i];
                now = i;
            
        
        if(now == -1) 
        
            cout << "impossible";
            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;

 

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