题意

题目链接

Sol

自己yy着写了一下Boruvka算法。

算法思想很简单，就是每次贪心的用两个联通块之间最小的边去合并。

复杂度\(O(n \log n)\)，然鹅没有Kruskal跑的快，但是好像在一类生成树问题上很有用

#include<bits/stdc++.h> 
#define Pair pair<int, int> 
#define fi first
#define se second
#define pb push_back
#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
char buf[(1 << 22)], *p1 = buf, *p2 = buf;
using namespace std;
const int MAXN = 5001;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, M;
namespace DSU {
    int fa[MAXN], siz[MAXN];
    void init(int N) {
        for(int i = 1; i <= N; i++) fa[i] = i, siz[i] = 1;
    }
    int find(int x) {
        return fa[x] == x ? fa[x] : fa[x] = find(fa[x]);
    }
    void unionn(int x, int y) {
        int fx = find(x), fy = find(y);
        if(siz[fx] > siz[fy]) swap(fx, fy);
        fa[fx] = fy; siz[fy] += siz[fx];
    }
};
using namespace DSU;
vector<Pair> v[MAXN];
int link[MAXN], val[MAXN];
void Boruvka() {
    init(N); int ans = 0;
    bool flag;
    do {
        flag = 0;
        memset(link, -1, sizeof(link));
        memset(val, 0x3f, sizeof(val));
        for(int x = 1; x <= N; x++) {
            int fx = find(x);
            for(auto &tmp : v[x]) {
                int to = tmp.fi, w = tmp.se, fy = find(to);
                if(fx == fy || (w > val[fx])) continue;
                link[fx] = fy; val[fx] = w;
            }
        }
        for(int x = 1; x <= N; x++) {
            int fx = find(x);
            if((~link[fx]) && find(fx) != find(link[fx])) 
                unionn(fx, link[fx]), ans += val[fx], flag = 1; 
        }   
    }while(flag);
    int f1 = find(1);
    for(int i = 2; i <= N; i++) if(find(i) != f1) return (void) puts("orz");
    cout << ans;
}
signed main() {
    N = read(); M = read();
    for(int i = 1; i <= M; i++) {
        int x = read(), y = read(), w = read();
        v[x].push_back({y, w});
        v[y].push_back({x, w});
    }
    Boruvka();
}