hdu 4661 Message Passing（树形dp）

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题目链接：hdu 4661 Message Passing

解题思路

考虑以每个位置为第一个汇总的点，即根节点。然后类似村名排队的计数。

代码

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;typedef long long ll;const int mod = 1e9 + 7;const int maxn = 1000000;int fac[maxn + 5], inv[maxn + 5];int pow_mod(ll x, int n) {    ll ret = 1;    while (n) {        if (n&1) ret = ret * x % mod;        x = x * x % mod;        n >>= 1;    }    return ret;}void presolve() {    fac[0] = 1;    for (int i = 1; i <= maxn; i++)        fac[i] = 1LL * i * fac[i-1] % mod;    inv[maxn] = pow_mod(fac[maxn], mod-2);    for (int i = maxn-1; i >= 0; i--)        inv[i] = 1LL * (i+1) * inv[i+1] % mod;}int N, ans, sz[maxn + 5];vector<int> G[maxn + 5];ll dp[maxn + 5];ll C(int n, int k) {    if (k > n || k < 0 || n < 0) return 0;    return 1LL * fac[n] * inv[k] % mod * inv[n-k] % mod;}ll dfs (int u, int f) {    ll& ret = dp[u];    ret = 1;    sz[u] = 0;    for (int i = 0; i < G[u].size(); i++) {        int v = G[u][i];        if (v == f) continue;        ret = ret * dfs(v, u) % mod;        sz[u] += sz[v];    }    int s = sz[u];    for (int i = 0; i < G[u].size(); i++) {        int v = G[u][i];        if (v == f) continue;        ret = ret * C(s, sz[v]) % mod;        s -= sz[v];    }    sz[u]++;    return ret;}void dfs(int u, int f, ll k) {    // k 为以u为根对应的反向树有多少种方式    ll ret = k;    k = (k * dp[u]) % mod;    int s = N-1;    for (int i = 0; i < G[u].size(); i++) {        int v = G[u][i];        if (v == f) continue;        // 扣除现在枚举的子树        ll iv = k * pow_mod(dp[v] * C(sz[u]-1, sz[v]) % mod, mod-2) % mod;        dfs(v, u, iv * C(N-sz[v]-1, sz[u]-sz[v]-1) % mod);        ret = ret * C(s, sz[v]) % mod * dp[v] % mod;        s -= sz[v];    }    ans = (ans + ret * ret % mod) % mod;}int main () {    presolve();    int cas;    scanf("%d", &cas);    while (cas--) {        scanf("%d", &N);        int u, v;        for (int i = 1; i <= N; i++) G[i].clear();        for (int i = 1; i < N; i++) {            scanf("%d%d", &u, &v);            G[u].push_back(v);            G[v].push_back(u);        }        ans = 0;        dfs(1, 0);        dfs(1, 0, 1);        printf("%d\n", ans);    }    return 0;}

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