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  • 牛客挑战赛30 小G砍树 树形dp

    小G砍树

    dfs两次, dp出每个点作为最后一个点的方案数。

    #include<bits/stdc++.h>
    #define LL long long
    #define fi first
    #define se second
    #define mk make_pair
    #define PLL pair<LL, LL>
    #define PLI pair<LL, int>
    #define PII pair<int, int>
    #define SZ(x) ((int)x.size())
    #define ull unsigned long long
    
    using namespace std;
    
    const int N = 1e5 + 7;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 998244353;
    const double eps = 1e-8;
    const double PI = acos(-1);
    
    int n, ans, son[N], dp[N];
    vector<int> G[N];
    
    int F[N], Finv[N], inv[N];
    
    void init() {
        inv[1] = F[0] = Finv[0] = 1;
        for(int i = 2; i < N; i++) inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
        for(int i = 1; i < N; i++) F[i] = 1ll * F[i - 1] * i % mod;
        for(int i = 1; i < N; i++) Finv[i] = 1ll * Finv[i - 1] * inv[i] % mod;
    }
    
    int Power(int a, int b) {
        int ans = 1;
        while(b) {
            if(b & 1) ans = 1ll * ans * a % mod;
            a = 1ll * a * a % mod; b >>= 1;
        }
        return ans;
    }
    
    void dfs(int u, int fa) {
        dp[u] = 1;
        for(auto& v : G[u]) {
            if(v == fa) continue;
            dfs(v, u);
            dp[u] = 1ll * dp[u] * dp[v] % mod;
            dp[u] = 1ll * dp[u] * Finv[son[v]] % mod;
            son[u] += son[v];
        }
        dp[u] = 1ll * dp[u] * F[son[u]] % mod;
        son[u]++;
    }
    
    void getAns(int u, int fa, int tmp) {
        ans = (ans + 1ll * tmp * dp[u] % mod * F[n - 1] % mod * Finv[son[u] - 1] % mod * Finv[n - son[u]] % mod) % mod;
        tmp = 1ll * tmp * Finv[n - son[u]] % mod;
        for(auto& v : G[u]) {
            if(v == fa) continue;
            tmp = 1ll * tmp * dp[v] % mod;
            tmp = 1ll * tmp * Finv[son[v]] % mod;
        }
        for(auto& v : G[u]) {
            if(v == fa) continue;
            int nxttmp = 1ll * tmp * F[son[v]] % mod * F[n - son[v] - 1] % mod * Power(dp[v], mod - 2) % mod;
            getAns(v, u, nxttmp);
        }
    }
    
    int main() {
        init();
        scanf("%d", &n);
        for(int i = 2; i <= n; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        dfs(1, 0);
        getAns(1, 0, 1);
        printf("%d
    ", ans);
        return 0;
    }
    
    /*
    */
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  • 原文地址:https://www.cnblogs.com/CJLHY/p/10509993.html
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