请问有没有大佬知道下面代码为什么使用连接一次边连通块数量减1就只能通过70分。
重新写个循环就可以AC。在印象中是可以这样判断有没有最小生成树把，而且同样的最小生成树H题这样写法都可以通过。

#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 ms(__x__,__val__) memset(__x__, __val__, sizeof(__x__))
#define rep(i, sta, en) for(int i=sta; i<=en; ++i)
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
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, int op = 10) { if (!x) { putchar('0'); if (op)    putchar(op); 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]);    if (op)    putchar(op); }
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; }
const int dir[][2] = { {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 7;
const int M = 2e6 + 7;

ll n, m;
int head[N], tot, fa[N];
map<string, int> mp;

struct Node {
    string u, v;
    ll w;
    bool operator < (const Node& opt) const {
        return w < opt.w;
    }
}edge[M];
int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }

void solve() {
    mp.clear();
    rep(i, 1, N - 1)    fa[i] = i;
    string st;
    cin >> st;
    int cnt = 0;
    rep(i, 1, m) {
        cin >> edge[i].u >> edge[i].v;
        cin >> edge[i].w;
        if (!mp.count(edge[i].u))
            mp[edge[i].u] = ++cnt;
        if (!mp.count(edge[i].v))
            mp[edge[i].v] = ++cnt;
    }
    sort(edge + 1, edge + 1 + m);
    cnt = n;
    ll ans = 0;
    rep(i, 1, m) {
        int fu = find(mp[edge[i].u]), fv = find(mp[edge[i].v]);
        if (fu != fv) {
            fa[fv] = fu;
            ans += edge[i].w;
            --cnt;
        }
        if (cnt == 1)    break;
    }
//     if (cnt == 1)    cout << ans << endl;  wa的
//     else cout << "NO!" << endl;
    int flag = 1; // ac的
    int flag1 = 1;
    rep(i, 1, n) {
        if (fa[i] == i && !flag) {
            flag1 = 0;
        }
        if (fa[i] == i && flag) {
            flag = 0;
        }
    }
    if (flag1)
        cout << ans << endl;
    else    cout << "No!" << endl;
}

int main() {
    js;
    //int T = read();
    while (cin >> n >> m)
        solve();
    return 0;
}