[2015HDU Multi-School league title]hdu5378 Leader in the Tree land

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5378

Test instructions: Give you a tree with a n-knot point. Because there is a root tree, each node can specify a subtree with its root (this is the subtree discussed later). Now we have assigned N minister from 1 to N to these nodes (one for each node), and the largest number of minister in any subtree is the leader of the subtree, asking how many different leaders you exist.

Solution:

Quote official:

The idea of probability can be used to solve this problem. Make the subtree with the node I as the subtrees tree, and set the subtree to have just sz[i] points. Then the first point is the probability of the maximum value of the subtrees tree is 1/sz[i], and the probability of not being the maximum is (sz[i]-1)/sz[i]. It is now possible to solve the probability that there are exactly k maximum values. DP[I][J] Represents a point that takes a number from 1 to I, where exactly J points are the probability of their subtree maximum. It is easy to get the following transfer equation: dp[i][j]=dp[i-1][j]* (sz[i]-1)/sz[i]+dp[i-1][j-1]/sz[i]. This dp[n][k] is the probability that all points have exactly the K-maximum value. The topic requires the number of programmes, with a total of n! The probability of multiplication is the answer. When calculating, replace the above score with the inverse element

1 /*2 * problem:hdu5378 Leader in the Tree land3 * AUTHOR:SHJWUDP4 * Created TIME:2015/8/12 Wednesday 20:17:315 * File name:1006.cpp6 * state:accepted7 * Memo: probability dp, multiplicative inverse8  */9#include <iostream>Ten#include <cstdio> One#include <vector> A#include <cstring> -#include <algorithm> -  the using namespacestd; -  - Const intmod=1e9+7; -  + structEdge { -     intu, v; + }; A  at intN, K; -Vector<edge>edges; -vector<vector<int> >G; -vector<int> siz, INV;///siz[i]*inv[i]=1 (mod 1e9+7) - voidinit () { - edges.clear (); inG.assign (n+1, vector<int> (0)); -Siz.resize (n+1); toInv.resize (n+1); + } - voidAddedge (intUintv) { the Edges.push_back (Edge) {u, v}); *G[u].push_back (Edges.size ()-1); $ }Panax Notoginseng intPow_mod (Long LongXintNintMoD) { -     Long Longres=1; the      while(n) { +         if(N &1) res=res*x%MoD; Ax=x*x%MoD; then>>=1; +     } -     returnRes; $ } $ voidDfsintUintFA) { -siz[u]=1; -      for(intI:g[u]) { theEdge & e=Edges[i]; -         if(E.V==FA)Continue;Wuyi dfs (E.V, u); thesiz[u]+=SIZ[E.V]; -     } Wu } - intMain () { About #ifndef Online_judge $Freopen ("inch","R", stdin); -     //freopen ("Out", "w", stdout); - #endif -     intT, now=0; Ascanf"%d", &T); +      while(t--) { thescanf"%d%d", &n, &k); - init (); $          for(intI=0; i<n-1; i++) { the             intA, B; thescanf"%d%d", &a, &b); the Addedge (A, b); the Addedge (b, a); -         } inDfs1, -1); the          for(intI=1; i<=n; i++) Inv[i]=pow_mod (Siz[i], mod-2, MOD);//multiplication inverse element for siz[i] thevector<vector<Long Long> > F (n+1, vector<Long Long> (n+1,0)); Aboutf[0][0]=1; the          for(intI=1; i<=n; i++) { the              for(intj=0; j<=k; J + +) { thef[i][j]+=f[i-1][j]* (siz[i]-1)%mod*inv[i]%MOD; +                 if(j>0) f[i][j]+=f[i-1][j-1]*inv[i]%MOD; -f[i][j]%=MOD; the             }Bayi         } the         Long Longans=F[n][k]; the          for(intI=1; i<=n; i++) { -ans= (ans*i)%MOD; -         } theprintf"Case #%d:", ++Now ); theprintf"%i64d\n", ans); the     } the     return 0; -}

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[2015HDU Multi-School league title]hdu5378 Leader in the Tree land