最小生成树 边权非负无向图

prim 算法

使用于领接矩阵版本
与dijkstra极其相似 只是更新的矩阵的数量不同

krukal算法

并查集
从小到大排序
一条边没有联通 就选择这条边
否则 如果联通就不选择这条边

求最大的边权最小

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int n,m;
const int N = 16005;
struct edge{
    int a,b,w;
    bool  operator<(edge &a){
        return w<a.w;
    }
}e[N];
int st[N];
int p[N];
int find(int x){
    if(x!=p[x]){
        p[x]=find(p[x]);
    }
    return p[x];
    
}
int main()
{
    cin >> n>>m;
    for (int i = 1; i <= n; i ++ ) p[i]=i;
    
    for (int i = 1; i <= m; i ++ ){
        cin >> e[i].a>>e[i].b>>e[i].w;
    }
    int res=0;
    sort(e+1,e+1+m);
    for (int i = 1; i <= m; i ++ ){
        int a=find(e[i].a),b=find(e[i].b);
        if(a!=b){
            p[a]=b;
            res=e[i].w;
        }
    }
    cout<<n-1<<" " << res;
    
    return 0;
}

必连边 缩点

再枚举非必边 https://www.acwing.com/activity/content/problem/content/1514/

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int n,m;
const int N = 10005;
int p[N];
struct edge{
    int a,b,w;
    
    bool  operator<(edge &a ){
        return w<a.w;
        
    }
}e[N];
int find(int x){
    if(x!=p[x]){
        p[x]=find(p[x]);
    }
    return p[x];
}
int main()
{
    cin >> n>>m;
    int res=0,k=0;
    for (int i = 1; i <= n; i ++ ) p[i]=i;
    for (int i = 1; i <= m; i ++ ){
        int t,a,b,c;
        cin >> t>>a>>b>>c;
        if(t==1){
            p[find(a)]=p[find(b)];
            res+=c;
        }
        else{
            e[k++]={a,b,c};
        }
    }
    
    sort(e,e+k);
    for (int i = 0; i < k; i ++ ){
        int a=find(e[i].a),b=find(e[i].b);
        if(a!=b){
            p[a]=b;
            res+=e[i].w;
        }
    }
    cout << res;
    return 0;
}

横竖的连线价值不一样 优先连接价值小的

https://www.acwing.com/problem/content/1146/

#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 1010, M = N * N, K = 2 * N * N;

int n, m, k;
int ids[N][N];
struct Edge
{
    int a, b, w;
}e[K];
int p[M];

int find(int x)
{
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
}

void get_edges()
{
    int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1}, dw[4] = {1, 2, 1, 2};

    for (int z = 0; z < 2; z ++ )//这个是方向 先竖再横
        for (int i = 1; i <= n; i ++ )//这个是坐标
            for (int j = 1; j <= m; j ++ )
                for (int u = 0; u < 4; u ++ )//枚举四个方向
                    if (u % 2 == z)//对2取模后刚好相等的就是对应的方向上的
                    {
                        int x = i + dx[u], y = j + dy[u], w = dw[u];
                        if (x && x <= n && y && y <= m)
                        {
                            int a = ids[i][j], b = ids[x][y];
                            if (a < b) e[k ++ ] = {a, b, w};//加一次就可以了 可以省点空间 但不加也可以啦
                        }
                    }
}

int main()
{
    cin >> n >> m;

    for (int i = 1, t = 1; i <= n; i ++ )
        for (int j = 1; j <= m; j ++, t ++ )
            ids[i][j] = t;

    for (int i = 1; i <= n * m; i ++ ) p[i] = i;

    int x1, y1, x2, y2;
    while (cin >> x1 >> y1 >> x2 >> y2)
    {
        int a = ids[x1][y1], b = ids[x2][y2];
        p[find(a)] = find(b);
    }

    get_edges();

    int res = 0;
    for (int i = 0; i < k; i ++ )
    {
        int a = find(e[i].a), b = find(e[i].b), w = e[i].w;
        if (a != b)
        {
            p[a] = b;
            res += w;
        }
    }

    cout << res << endl;

    return 0;
}

有点权 有边权

https://www.acwing.com/problem/content/1148/
建立虚拟源点 将点权变成边权 转化为最小生成树

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 305;
int w[N][N];
int n;
int d[N];
bool st[N];
int p(){
    memset(d, 0x3f, sizeof d);
    d[0]=0;
    int res=0;
    for (int i = 0; i <= n; i ++ ){
        int t=-1;
        
        for (int j = 0; j <= n; j ++ ){
            if(!st[j]&&(t==-1||d[t]>d[j])){
                t=j;
                
            }
        }
        st[t]=true;
        res+=d[t];
        
        for (int j = 0; j <= n; j ++ ){
            d[j]=min(d[j],w[t][j]);    
        }
    }
    
    
    
    return res;
}
int main()
{
    cin >> n;
    for (int i = 1; i <= n; i ++ ){
        cin >> w[0][i];
        w[i][0]=w[0][i];
    }
    for (int i = 1; i <= n; i ++ ){
        for (int j = 1; j <= n; j ++ ){
            cin >> w[i][j];
        }
    }
    cout << p();
    return 0;
}