AcWing859.Kruskal算法求最小生成树

题解

可以通过并查集查看a,b的根结点是否相同，相同则代表连通，即会成环不能加入最小生成树

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

using namespace std;

const int N = 1e5 + 10, M = 2e5 + 10;

struct Edge{
    int a, b, c;
    bool operator<(const Edge &t){
        return c < t.c;
    }
}edge[M];

int p[N], n, m;

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

int Kruskal()
{
    int pa, pb;
    int res = 0, cnt = 0;
    sort(edge, edge+m);
    for(int i = 0; i < m; ++i)
    {
        pa = find(edge[i].a), pb = find(edge[i].b);
        if(pa != pb) //不连通即不成坏
            res += edge[i].c, p[pa] = pb, cnt++;
    }
    if(cnt < n-1) return 0x3f3f3f3f;
    return res;
}

int main()
{
    scanf("%d%d",&n, &m);
    
    for(int i = 1; i <= n; ++i)
        p[i] = i;
    
    int a, b, c;
    for(int i = 0; i < m; ++i)
    {
        scanf("%d%d%d",&a, &b, &c);
        edge[i] = {a, b, c};
    }
    int res = Kruskal();
    if(res == 0x3f3f3f3f) cout << "impossible" << endl;
    else cout << res << endl;
    return 0;
}