局域网——Prim Kruskal

题目

Prim （生成一颗包含起点的最小生成树，所以要多次调用）

cpp 复制代码

#include <bits/stdc++.h>

using namespace std;

const int N = 510;
const int inf = 0x3f3f3f3f;

int n, m;
int g[N][N], dis[N];
bool p[N], vis[N];

int prim (int u)
{
    memset(dis, 0x3f, sizeof dis); dis[u] = 0;
    int sum = 0;
    for(int i = 0 ; i < n ; i ++ )
    {
        int t = -1;
        for(int j = 1 ; j <= n ; j ++ )
            if(!p[j] && (t == -1 || dis[t] > dis[j]))
                t = j;
        if(i && dis[t] == inf) return sum;
        p[t] = 1;
        if(i) sum += dis[t]; // 第一个点为根节点没有边权
        vis[t] = 1;
        for(int j = 1 ; j <= n ; j ++ )
            if(!p[j] && dis[j] > g[t][j]) dis[j] = g[t][j];
    }
    return sum;
}

int main ()
{
    int ans = 0;
    cin >> n >> m;
    for(int i = 1 ; i <= n ; i ++ )
        for(int j = 1 ; j <= n ; j ++ )
            if(i == j) g[i][j] = 0;
            else g[i][j] = inf;
    for(int i = 1 ; i <= m ; i ++ )
    {
        int a, b, c;
        cin >> a >> b >> c;
        g[a][b] = g[b][a] = min(g[a][b], c);
        ans += min(g[a][b], c);
    }
    int t = 0;
    for(int i = 1 ; i <= n ; i ++ )
        if(!vis[i]) t += prim(i);
    cout << ans - t << endl;
    return 0;
}

Kruskal （如果有多颗生成树，生成最小生成森林）

cpp 复制代码

#include <bits/stdc++.h>
using namespace std;
const int N = 110;
const int M = 210;
struct edge{
    int a;
    int b;
    int c;
    
    bool operator < (const edge& v)
    {
        return c < v.c;
    }
} e[M];
int p[N];
int n, idx, m;
int find(int x)
{
    if(p[x] != x) p[x] = find(p[x]);
    return p[x];
}
int kruskal()
{
    int retv = 0;
    
    for(int i = 1; i <= n; i++)
        p[i] = i;
    sort(e+1,e+m+1);
    
    for(int i = 1; i <= m; i++)
    {
        int a = e[i].a, b = e[i].b, c = e[i].c;
        a = find(a), b = find(b);
        if(a != b)
        {
            p[a] = b;
            retv += c;
        }
    }
    
    return retv;
}
int main()
{
    cin >> n >> m;
    int sum = 0;
    for(int i = 1; i <= m; i++)
    {
        int a, b, c;
        cin >> a >> b >> c;
        e[++idx]  = {a, b, c};
        sum += c;
    }
            
    int t = kruskal();
    cout << sum - t;
}