Description

给定一棵\(n\)个点的树和正整数\(k\)，每条边长度都为\(1\)，对于每个点\(u\)求：\(\sum_{j=1}^{n}dist(u,j)^k;\) \(n<=50000,k<=150\)，答案对\(10007\)取模

Solution

组合数学+树形\(dp\)

先给出一个式子：
\(a^k=\sum_{i=1}^{k}S(k,i)*i!*C(a,i)\)
解释：\(a^k\)看作\(k\)个球放入\(a\)个有区别的盒子里的方案数
\(S(k,i)\)为第二类斯特林数，表示\(k\)个球放入\(i\)个盒子，不允许盒子空着，且这\(i\)个盒子无区别的方案数
\(i!*C(a,i)\)其实就是\(A(a,i)\)
上式即枚举\(i\)个盒子非空，用\(S(k,i)\)乘上a个盒子中，有序选i个的方案数
因此原式 \(=\sum_{j=1}^{n}\sum_{i=1}^{k}S(k,i)*i!*C(dist(u,j),i)\)
考虑先枚举\(i\)，那么原式化为\(\sum_{i=1}^{k}S(k,i)*i!*\sum_{j=1}^{n}C(dist(u,j),i)\)
其中\(S(k,i)=S(k-1,i)*i+S(k-1,i-1)\)
问题转化为求\(\sum_{j=1}^{n}C(dist(u,j),i)\)
考虑树形\(dp\)，即将\(j\)按是否在\(u\)子树内分类
设\(f[u][i]\)表示\(\sum_{j在u子树内}C(dist(u,j),i)\)
众所周知\(C(x,y)=C(x-1,y-1)+C(x-1,y)\)
于是枚举\(u\)的每个儿子\(v\)进行递推：
\(f[u][i]+=\sum_{j在v子树内}C(dist(v,j),i)+C(dist(v,j),i-1)\)
即\(f[u][i]+=f[v][i]+f[v][i-1]\)
注意特判\(f[u][0]+=f[v][0]\)，即\(i=0\)的时候不要\(+=f[v][i-1]\)
设\(g[u][i]\)表示\(\sum_{j=1}^{n}C(dist(u,j),i)\)
枚举\(u\)的每个儿子\(v\)
能给\(g[v][i]\)贡献的部分即\(\sum_{j不在v子树中}C(dist(u,j),i)\) ，记为\(now[i]\)
显然\(now[i]=g[u][i]-f[v][i-1]-f[v][i]\)
再递推到\(v\)即\(g[v][i]=f[v][i]+now[i]+now[i-1]\)
与上文\(f[u][0]\)同理，注意特判\(g[v][0]\)

Code

#include <bits/stdc++.h>

using namespace std;

template <class t>
inline void read(t & res)
{
   char ch;
   while (ch = getchar(), !isdigit(ch));
   res = ch ^ 48;
   while (ch = getchar(), isdigit(ch))
   res = res * 10 + (ch ^ 48);
}

const int e = 50005, o = 155, mod = 10007;
int n, m, f[e][o], g[e][o], now[o], s[o][o], adj[e], nxt[e * 2], go[e * 2], num, fac[o];

inline void link(int x, int y)
{
   nxt[++num] = adj[x];
   adj[x] = num;
   go[num] = y;
   nxt[++num] = adj[y];
   adj[y] = num;
   go[num] = x; 
}

inline void add(int &x, int y)
{
   x += y;
   while (x >= mod) x -= mod;
}

inline void dfs1(int u, int pa)
{
   int i, j;
   f[u][0] = 1;
   for (i = adj[u]; i; i = nxt[i])
   {
       int v = go[i];
       if (v == pa) continue;
       dfs1(v, u);
       add(f[u][0], f[v][0]);
       for (j = 1; j <= m; j++) add(f[u][j], f[v][j - 1] + f[v][j]);
   }
   for (j = 0; j <= m; j++) g[u][j] = f[u][j];
}

inline void dfs2(int u, int pa)
{
   int i, j;
   for (i = adj[u]; i; i = nxt[i])
   {
       int v = go[i];
       if (v == pa) continue;
       for (j = 0; j <= m; j++) now[j] = g[u][j];
       add(now[0], mod - f[v][0]);
       for (j = 1; j <= m; j++) add(now[j], 2 * mod - f[v][j] - f[v][j - 1]);
       add(g[v][0], now[0]);
       for (j = 1; j <= m; j++) add(g[v][j], now[j - 1] + now[j]);
       dfs2(v, u);
   }
}

int main()
{
   int i, j, x, y;
   read(n); read(m);
   for (i = 1; i < n; i++)
   {
       read(x);
       read(y);
       link(x, y);
   }
   fac[0] = 1;
   for (i = 1; i <= m; i++) fac[i] = fac[i - 1] * i % mod;
   for (i = 1; i <= m; i++)
   for (j = 1; j <= i; j++)
   if (j == 1) s[i][j] = 1;
   else s[i][j] = (s[i - 1][j] * j + s[i - 1][j - 1]) % mod;
   dfs1(1, 0);
   dfs2(1, 0);
   for (i = 1; i <= n; i++)
   {
       int ans = 0;
       for (j = 1; j <= m; j++) add(ans, 1ll * s[m][j] * fac[j] * g[i][j] % mod);
       printf("%d\n", ans);
   }
   return 0;
}

[BZOJ2159][洛谷P4827][国家集训队] Crash 的文明世界（第二类斯特林数+dp）

Description

Solution

Code

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