UVA 11174 - Stand in a Line (数学基础+除法取模)

Problem J
Stand in a Line
Input: Standard Input

Output: Standard Output

 

All the people in the byteland want to stand in a line in such a way that noperson stands closer to the front of the line than his father. You are giventhe information about the people of the byteland.  You have to determine the number of ways the bytelandian people can stand in a line.

 

Input

First line of the input containsT (T<14) the number of test case. Then following lines contains T Testcases.

Each test case starts with 2integers n (1≤n≤40000) and m (0≤m<n).  n is the number ofpeople in the byteland and m is the number of peoplewhose father is alive. These n people are numbered 1...n.Next m line contains two integers aand b denoting that b is the father of a. Each person can have at mostone father. And no person will be an ancestor of himself.

 

 

Output

For each test case the outputcontains a single line denoting the number of different ways the soldier canstand in a single line. The result may be too big. So always output the remainder on dividing ther the result by 1000000007.

 

Sample Input                                  Output forSample Input

3

3 2

2 1

3 1

3 0

3 1

2 1

2

6

3

 

 

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Problemsetter: Abdullah-al-Mahmud

Special Thanks: Derek Kisman

 

 

村名排队问题。n个人排成一列，没有人站在他父亲前面的方法数。

这一题主要还是组合数学的问题，

公式就是，n!/(cnt[1]*cnt[2]*....cnt[n) cnt[i]为i的子树的大小

用到除法求模，求逆元

#include <iostream>#include <cstdio>#include <vector>using namespace std;const int mod=1000000007;const int maxn=40000;int n,m,cnt[maxn+10];long long fact[maxn+10];vector <vector<int> > v;void ini(){    fact[0]=1,fact[1]=1;    for(int i=2;i<=maxn;i++){        fact[i]=(fact[i-1]*i)%mod;    }}void initial(){    for(int i=0;i<=n;i++){        cnt[i]=-1;    }    v.clear();}void input(){    int x,y;    v.resize(n+1);    while(m-- >0){        scanf("%d%d",&x,&y);        v[y].push_back(x);    }}int dfs(int x){    if(cnt[x]!=-1) return cnt[x];    int c=1;    for(int i=0;i<v[x].size();i++){        cnt[v[x][i]]=dfs(v[x][i]);        c+=cnt[v[x][i]];    }    return c;}long long pow_mod(long long a,long long p,long long n){    if(p==0) return 1;    long long ans=pow_mod(a,p/2,n);    ans=ans*ans%n;    if(p%2==1) ans=ans*a%n;    return ans;}void computing(){    for(int i=1;i<=n;i++){        cnt[i]=dfs(i);    }    long long ans=fact[n];    for(int i=1;i<=n;i++){        ans=( ans*pow_mod(cnt[i],mod-2,mod))%mod;    }    cout<<ans<<endl;}int main(){    //freopen("in","r",stdin);    ini();    int t;    scanf("%d",&t);    while(t-- >0){        scanf("%d%d",&n,&m);        initial();        input();        computing();    }    return 0;}