[Probability DP] HDU 5378 Leader in the Tree land

Test instructions

Give you a tree with a 1-bit root node, we define for each subtree, the weights of the node weights are recorded as the weights of this subtree, for you to put 1~n into this tree

The number of combinations that satisfy the maximum weight is only K.

Ideas:

We can know what the probability is that each node is a subtree, and the maximum value of the root node is the number of probabilities.

Then actually the problem becomes we in n items, each item takes the probability is pi does not take the probability is 1-pi

What's the probability of you getting k items?

And then the last ride n! Just fine. The intermediate computation uses the inverse element.

Code:

#include "cstdlib" #include "CString" #include "Cmath" #include "Cstdio" #include "queue" #include "algorithm" #include " Iostream "#include" stack "using namespace std; #define LL __int64#define N 123456vector<int>edge[2234];ll sum[1234    ],dp[1234][1234];ll In[1234],mod=1000000007ll;ll Power (ll A,ll b) {ll ans=1;        while (b) {if (b&1) ans= (ans*a)%mod;        A= (a*a)%mod;    b>>=1; } return ans;    void Dfs (int x,int f) {int ans=1,lit=edge[x].size ();        for (int i=0; i<lit; i++) {int v=edge[x][i];        if (v==f) continue;        DFS (V,X);    ANS+=SUM[V];    } Sum[x]=ans; return;}    int main () {int t,cas=1;    cin>>t; for (int i=1; i<=1234; i++) In[i]=power (i,mod-2);        preprocessing inverse while (t--) {int n,k;        scanf ("%d%d", &n,&k);        for (int i=1; i<=n; i++) edge[i].clear ();            for (int i=1; i<n; i++) {int x, y;            scanf ("%d%d", &x,&y);      Edge[x].push_back (y);      Edge[y].push_back (x);        } memset (Sum,0,sizeof (sum));        DFS (a);        Memset (Dp,0,sizeof (DP));        Dp[0][0]=1; for (int i=1, i<=n; i++) {for (int j=0; j<=k; J + +) {if (j>0) dp[i][j]=                (Dp[i][j]+dp[i-1][j-1]*in[sum[i]])%mod;            Dp[i][j]= (dp[i][j]+ (dp[i-1][j]* (sum[i]-1)%mod) *in[sum[i]]%mod)%mod;        }} ll Ans=1;        for (int i=1; i<=n; i++) ans= (ans*i)%mod;        ANS*=DP[N][K];    printf ("Case #%d:%i64d\n", cas++,ans%mod); } return 0;}

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[Probability DP] HDU 5378 Leader in the Tree land