这个题真的是有点恶心了。
ans[ i ] 表示以 i 为根的子树的答案, 也就是所有合法排列的逆序对数量。
way[ i ] 表示以 i 为根的合法排列数量。
dp[ i ][ j ][ k ] 表示以 i 为根所有合法排列中 j 节点排在第 k 个的合法排列数量。
然后用这三个dp 数组去更新答案。
具体参考https://blog.csdn.net/ShinFeb/article/details/52212775
#pragma GCC optimize(2) #pragma GCC optimize(3) #include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned 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 ALL(x) (x).begin(), (x).end() #define fio ios::sync_with_stdio(false); cin.tie(0); using namespace std; const int N = 3e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = (int)1e9 + 7; const double eps = 1e-8; const double PI = acos(-1); template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;} template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;} template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;} template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;} int n, root; int ans[51], way[51], sz[51]; int dp[51][51][51]; int tmp[51][51]; int sum[51][51]; int C[101][101]; vector<int> G[N]; void dfs(int u, int fa) { for(int i = 0; i < SZ(G[u]); i++) { if(G[u][i] == fa) { G[u].erase(G[u].begin() + i); } } ans[u] = 0; sz[u] = 1; for(auto &v : G[u]) { dfs(v, u); sz[u] += sz[v]; } for(int i = 1; i <= n; i++) { for(int j = 1; j <= n; j++) { dp[u][i][j] = 0; sum[i][j] = 0; } } way[u] = 1; int cur_cnt = 0; for(auto &v : G[u]) { int nex_cnt = cur_cnt + sz[v]; ans[u] = 1LL * C[nex_cnt][cur_cnt] * ans[u] % mod * way[v] % mod; add(ans[u], 1LL * C[nex_cnt][cur_cnt] * ans[v] % mod * way[u] % mod); for(int i = 1; i <= n; i++) { for(int k = 1; k <= sz[v]; k++) { if(!dp[v][i][k]) continue; for(int j = 0; j <= cur_cnt; j++) { int a = 1LL * C[k + j - 1][j] * C[cur_cnt - j + sz[v] - k][cur_cnt - j] % mod; int b = (sum[i][cur_cnt] - sum[i][j] + mod) % mod; int c = (sum[n][j] - sum[i][j] + mod) % mod; add(ans[u], 1LL * dp[v][i][k] * a % mod * (b + c) % mod); } } } for(int i = 1; i <= n; i++) { for(int j = 1; j <= n; j++) { tmp[i][j] = 0; } } for(int i = 1; i <= n; i++) { for(int j = 1; j <= nex_cnt; j++) { for(int k = 1; k <= j; k++) { int a = 1LL * C[j - 1][k - 1] * C[cur_cnt + sz[v] - j][cur_cnt - k] % mod; int b = 1LL * C[j - 1][k - 1] * C[cur_cnt + sz[v] - j][sz[v] - k] % mod; add(tmp[i][j], 1LL * a * way[v] % mod * dp[u][i][k] % mod); add(tmp[i][j], 1LL * b * way[u] % mod * dp[v][i][k] % mod); } } } for(int i = 1; i <= n; i++) { for(int j = 1; j <= n; j++) { dp[u][i][j] = tmp[i][j]; sum[i][j] = dp[u][i][j]; add(sum[i][j], sum[i - 1][j]); add(sum[i][j], sum[i][j - 1]); sub(sum[i][j], sum[i - 1][j - 1]); } } way[u] = 1LL * way[u] * way[v] % mod * C[nex_cnt][cur_cnt] % mod; cur_cnt = nex_cnt; } dp[u][u][sz[u]] = way[u]; add(ans[u], sum[n][cur_cnt]); sub(ans[u], sum[u][cur_cnt]); } void init() { for(int i = 1; i <= n; i++) { G[i].clear(); } } int main() { for(int i = 0; i <= 100; i++) { for(int j = C[i][0] = 1; j <= i; j++) { C[i][j] = C[i - 1][j - 1] + C[i - 1][j]; if(C[i][j] >= mod) C[i][j] -= mod; } } while(scanf("%d%d", &n, &root) != EOF) { init(); for(int i = 1; i < n; i++) { int u, v; scanf("%d%d", &u, &v); G[u].push_back(v); G[v].push_back(u); } dfs(root, 0); printf("%d ", ans[root]); } return 0; } /* */