Uva 5002-the Queue DFS

On some special occasions Nadia ' s company provide very special lunch for all employees of the company. Before the food was served all of the employees must stand in a queue with front of the food counter. The company applied a rule for standing in the queue. The rule is nobody can stand anywhere in front of its supervisor in the queue. For example if Abul are the supervisor of Babul and Abul stands in kth position from the front of the queue, then Babul can Not stand at no position in between 1 and k–1 from front of the queue.

The company had N employees and each of them had exactly one supervisor except one who doesn ' t had any superviso R.

You have to calculate in how many ways the queue can be created. For this problem, you can safely assume, and the the queue can be created on least one.

Input

Input starts with a integer t (t is around ), the number of test cases.

Each test case is starts with a line containing one integer N (1 ≤n≤1000). Each of the following N-1 lines would contain the integers a and B (1 ≤ A, b ≤ N and a ≠ b), which denotes a is the supervisor of B. For the sake of simplicity we is representing each employee by an integer number.

Output

For each of the input case, the output a of the format "Case #: W", Here's the case number and W is The number of ways to create the queue. The number of ways can very large. You have to print the number modulo 1,000,000,007.

Sample input Output for sample input

1 5 2 1 2 3 3 4 3 5

Case 1:8

#include <cstdio>#include<cmath>#include<cstring>#include<ctime>#include<iostream>#include<algorithm>#include<Set>#include<vector>#include<sstream>#include<queue>#include<typeinfo>#include<fstream>typedefLong Longll;using namespacestd;//freopen ("d.in", "R", stdin);//freopen ("D.out", "w", stdout);#defineSspeed ios_base::sync_with_stdio (0); Cin.tie (0)#defineMAXN 1010#defineMOD 1000000007Const intinf=0x7fffffff;//infinitely Largell yh[ .][ .];voidBuildyanghui (ll N) {ll i,j; yh[0][0]=1; yh[0][1]=0;  for(i=1; i<=n;i++) {yh[i][0]=1;  for(j=1; j<=n;j++) {Yh[i][j]= (yh[i-1][j-1]+yh[i-1][J])%MOD; }    }}structnode{intPre;    ll num; Vector<ll>God; intans;}; Node KILL[MAXN];voidDfsintN) {    if(kill[n].god.size () = =0) {Kill[n].num=0; return; }    Else    {         for(intI=0; I<kill[n].god.size (); i++)        {            if(kill[kill[n].god[i]].num==0) DFS (Kill[n].god[i]); Kill[n].num+=kill[kill[n].god[i]].num+1; }    }}voidDFS1 (ll N) {if(kill[n].god.size () = =0) {Kill[n].ans=1; return; } Kill[n].ans=1; if(kill[n].god.size () = =1)    {        if(kill[kill[n].god[0]].ans==0) {DFS1 (kill[n].god[0]); } Kill[n].ans=kill[kill[n].god[0]].ans; return; } ll Num=0;  for(intI=0; I<kill[n].god.size (); i++)    {        if(kill[kill[n].god[i]].ans==0) {DFS1 (kill[n].god[i]); }        if(i==0) {num=kill[kill[n].god[i]].num+1; Kill[n].ans=kill[n].ans*kill[kill[n].god[i]].ans%MOD; Kill[n].ans%MOD; Continue; } Kill[n].ans=kill[n].ans*yh[num+kill[kill[n].god[i]].num+1][kill[kill[n].god[i]].num+1]%mod*kill[kill[n].god[i]].ans%MOD; Kill[n].ans%=MOD; Num+=kill[kill[n].god[i]].num+1; }}intMain () {intT; CIN>>T; Buildyanghui (2001);  for(intcas=1; cas<=t;cas++) {memset (Kill,0,sizeof(Kill)); intN; CIN>>N;        ll A, B;  for(intI=0; i<n-1; i++) {cin>>a>>b; Kill[b].pre=A;        Kill[a].god.push_back (b); }        intSB;  for(intI=1; i<=n;i++)        {            if(kill[i].pre==0) SB=i;        } dfs (SB);        DFS1 (SB); cout<<" Case"<<cas<<":"<<" "<<kill[sb].ans%MOD<<Endl; }    return 0;}

Uva 5002-the Queue DFS