最小生成树管道铺设c++代码

根据提供的引用内容，以下是使用C++实现Prim算法和Kruskal算法求解最小生成树的代码：

1. Prim算法

#include <iostream>
#include <cstring>
using namespace std;

const int N = 510;
int g[N][N], dist[N];
bool st[N];

int prim(int n)
{
    memset(dist, 0x3f, sizeof dist);
    int res = 0;
    for (int i = 0; i < n; i++)
    {
        int t = -1;
        for (int j = 1; j <= n; j++)
            if (!st[j] && (t == -1 || dist[t] > dist[j]))
                t = j;
        if (i && dist[t] == 0x3f3f3f3f) return -1;
        if (i) res += dist[t];
        st[t] = true;
        for (int j = 1; j <= n; j++) dist[j] = min(dist[j], g[t][j]);
    }
    return res;
}

int main()
{
    int n, m;
    cin >> n >> m;
    memset(g, 0x3f, sizeof g);
    while (m--)
    {
        int a, b, c;
        cin >> a >> b >> c;
        g[a][b] = g[b][a] = min(g[a][b], c);
    }
    int t = prim(n);
    if (t == -1) puts("impossible");
    else cout << t << endl;
    return 0;
}

2. Kruskal算法

#include <iostream>
#include <algorithm>
using namespace std;

const int N = 200010;
int p[N];

struct Edge
{
    int a, b, w;
    bool operator< (const Edge &W) const
    {
        return w < W.w;
    }
}edges[N];

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

int main()
{
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < m; i++)
        cin >> edges[i].a >> edges[i].b >> edges[i].w;
    sort(edges, edges + m);
    for (int i = 1; i <= n; i++) p[i] = i;
    int res = 0, cnt = 0;
    for (int i = 0; i < m; i++)
    {
        int a = edges[i].a, b = edges[i].b, w = edges[i].w;
        a = find(a), b = find(b);
        if (a != b)
        {
            p[a] = b;
            res += w;
            cnt++;
        }
    }
    if (cnt < n - 1) puts("impossible");
    else cout << res << endl;
    return 0;
}