题目意思

给出n个点m条边，找出最小生成数的花费

解题思路

挺简单的最多100000个点，500000条边，可以看到是个稀疏图大概，采用kruskal求最小生成树

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define pai pair<int, int>
#define mk(__x__,__y__) make_pair(__x__,__y__)
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const int MOD = 1e9 + 7;
const ll INF = 0x3f3f3f3f;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void print(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 1e5 + 7;
const int M = 5e5 + 7;
struct Node {
    int u, v, w;
    bool operator < (const Node& tmp) const {
        return w < tmp.w;
    }
}edge[M << 1];
int fa[N];

int find(int x) {
    return x == fa[x] ? x : fa[x] = find(fa[x]);
}

int main() {
    int n = read(), m = read();
    for (int i = 1; i <= n; ++i)    fa[i] = i;
    for (int i = 1; i <= m; ++i) {
        int u = read(), v = read(), w = read();
        edge[i] = { u,v,w };
    }
    sort(edge + 1, edge + 1 + m);
    ll ans = 0;
    for (int i = 1; i <= m; ++i) {
        int fu = find(edge[i].u), fv = find(edge[i].v);
        if (fu != fv)    fa[fu] = fv, ans += edge[i].w;
    }
    print(ans);
    return 0;
}