子树的决策问题一般都是树形, 定义状态表示表示度数为的点无法到达的最小代价

  • 情况, 切断当前边代价为
  • 情况, 切断子树的某条边代价为 两种取最小值即可
#include <bits/stdc++.h>

#define x first
#define y second
#define all(x) x.begin(), x.end()

using namespace std;
using i128 = __int128;
using u128 = unsigned __int128;

typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;

const int N = 1e5 + 10, MOD = 998244353;
const int INF = 1e9;
const LL LL_INF = 1e18;
const LD EPS = 1e-8;
const int dx4[] = {-1, 0, 1, 0}, dy4[] = {0, 1, 0, -1};
const int dx8[] = {-1, -1, -1, 0, 0, 1, 1, 1}, dy8[] = {-1, 0, 1, -1, 1, -1, 0, 1};

istream &operator>>(istream &is, i128 &val) {
    string str;
    is >> str;
    val = 0;
    bool flag = false;
    if (str[0] == '-') flag = true, str = str.substr(1);
    for (char &c: str) val = val * 10 + c - '0';
    if (flag) val = -val;
    return is;
}

ostream &operator<<(ostream &os, i128 val) {
    if (val < 0) os << "-", val = -val;
    if (val > 9) os << val / 10;
    os << static_cast<char>(val % 10 + '0');
    return os;
}

void solve() {
    int n, m, s;
    cin >> n >> m >> s;
    vector<vector<PII>> g(n + 1);
    for (int i = 0; i < m; ++i) {
        int a, b, w;
        cin >> a >> b >> w;
        g[a].push_back({b, w});
        g[b].push_back({a, w});
    }

    vector<LL> f(n + 1);
    function<void(int, int)> dfs = [&](int u, int fa) {
        for (auto [v, w] : g[u]) {
            if (v == fa) continue;
            dfs(v, u);
            if (g[v].size() == 1) f[u] += w;
            else f[u] += min(1ll * w, f[v]);
        }
    };

    dfs(s, 0);

    cout << f[s] << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int T = 1;
    while (T--) solve();
    cout << fixed << setprecision(15);

    return 0;
}