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最新推荐文章于 2025-02-15 22:33:25 发布

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最新推荐文章于 2025-02-15 22:33:25 发布

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本文链接：https://blog.youkuaiyun.com/a_siscon/article/details/8764844

本文介绍了一种基于图论和组合数学的算法解决方案，通过预计算组合数、使用递归深度优先搜索等方法来解决特定的图论问题，如计算一棵树的不同排列数量。

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https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3003

# include <cstdio>
# include <iostream>
# include <set>
# include <map>
# include <vector>
# include <list>
# include <queue>
# include <stack>
# include <cstring>
# include <string>
# include <cstdlib>
# include <cmath>
# include <algorithm>

using namespace std ;

const int mod = 1e9 + 7 ;
long long c [ 1100 ] [ 1100 ] ;

void getC ( )
{
    for ( int i = 1 ; i <= 1000 ; i ++ )
        c [ i ] [ 0 ] = 1 , c [ i ] [ i ] = 1 ;
    for ( int j = 1 ; j <= 1000 ; j ++ )
    {
        for ( int i = j + 1 ; i <= 1000 ; i ++ )
            c [ i ] [ j ] = ( c [ i - 1 ] [ j ] + c [ i - 1 ] [ j - 1 ] ) % mod ;
    }
}

bool froot [ 1100 ] ;

struct Edge
{
    int v , next ;
} edge [ 3000 ] ;
int head [ 1100 ] ;
int cnt ;

void addedge ( int u , int v )
{
    edge [ cnt ] . v = v ;
    edge [ cnt ] . next = head [ u ] ;
    head [ u ] = cnt ++ ;
}

bool vis [ 1100 ] ;
int num [ 1100 ] ;

int getnum ( int u )
{
    vis [ u ] = 1 ;
    for ( int i = head [ u ] ; i != -1 ; i = edge [ i ] . next )
    {
        int v = edge [ i ] . v ;
        if ( ! vis [ v ] )
        {
            num [ u ] += getnum ( v ) ;
        }
    }
    return num [ u ] + 1 ;
}

long long per [ 1100 ] ;

long long dfs ( int u )
{
    vis [ u ] = 1 ;
    int sum = num [ u ] ;
    for ( int i = head [ u ] ; i != -1 ; i = edge [ i ] . next )
    {
        int v = edge [ i ] . v ;
        if ( ! vis [ v ] )
        {
            per [ u ] = ( ( per [ u ] * dfs ( v ) ) % mod * c [ sum ] [ num [ v ] + 1 ] ) % mod ;
            sum = sum - num [ v ] - 1 ;
        }
    }
    return per [ u ] ;
}

int main ( )
{
    getC ( ) ;
    int t ;
    cin >> t ;
    for ( int z = 1 ; z <= t ; z ++ )
    {
        cnt = 0 ;
        memset ( head , -1 , sizeof ( head ) ) ;
        memset ( froot , 0 , sizeof ( froot ) ) ;
        memset ( vis , 0 , sizeof ( vis ) ) ;
        memset ( num , 0 , sizeof ( num ) ) ;
        int n ;
        scanf ( "%d" , & n ) ;
        for ( int i = 1 ; i < n ; i ++ )
        {
            int u , v ;
            scanf ( "%d%d" , & u , & v ) ;
            froot [ v ] = 1 ;
            addedge ( u , v ) ;
        }
        int root = 0 ;
        for ( int i = 1 ; i <= n ; i ++ )
            if ( ! froot [ i ] )
            {
                root = i ;
                break ;
            }
        getnum ( root ) ;
        for ( int i = 1; i <= n ; i ++ )
            per [ i ] = 1 ;
        memset ( vis , 0 , sizeof ( vis ) ) ;
        cout << "Case " << z << ": " ;
        cout << dfs ( root ) << endl ;
    }
}