【AGC 002F】Leftmost Ball

Description

Snuke loves colorful balls. He has a total of N*K balls, K in each of his favorite N colors. The colors are numbered 1 through N.He will arrange all of the balls in a row from left to right, in arbitrary order. Then, for each of the N colors, he will paint the leftmost ball of that color into color 0, a color different from any of the N original colors.After painting, how many sequences of the colors of the balls are possible? Find this number modulo 10⁹+7. 1≤N,K≤2000

Input

The input is given from Standard Input in the following format: N K

Output

Print the number of the possible sequences of the colors of the balls after painting, modulo 10⁹+7.

 

题意：有$n$种颜色的球，标号$1$到$n$，每种颜色有$k$个。将$nk$个球随机排列后，将每种颜色的第一个球涂成颜色$0$，求最终可能得到的颜色序列的方案数。

分析：

令$f(i,j)~(i\leq j)$表示已经放置了i个编号为0的球与j种第一次出现的位置最靠前的颜色的方案数。每次在当前的第一个空位放置一个颜色为$0$的球或是一种未出现的颜色的球。可得转移方程：

$$f(i,j)=f(i-1,j)+\binom{n-i+(n-j+1)\cdot(k-1)-1}{k-2}\cdot(n-j+1)\cdot f(i,j-1)$$

时间复杂度$O(nk)$。

 

#include<cstdio>
#include<algorithm> 
#include<cstring>
#define LL long long
using namespace std;
const int N=2e3+5;
const int mod=1e9+7;
int n,m,fac[N*N],inv[N*N],f[N][N];
int read()
{
    int x=0,f=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
void Mod(int& a,int b){a+=b;if(a>=mod)a-=mod;}
int power(int a,int b)
{
    int ans=1;
    while(b)
    {
        if(b&1)ans=1ll*ans*a%mod;
        a=1ll*a*a%mod;b>>=1;
    }
    return ans;
}
int C(int n,int m){return 1ll*fac[n]*inv[m]%mod*inv[n-m]%mod;}
int main()
{
    n=read();m=read();
    if(m==1){printf("1");return 0;}
    fac[0]=1;
    for(int i=1;i<=n*m;i++)fac[i]=1ll*fac[i-1]*i%mod;
    inv[n*m]=power(fac[n*m],mod-2);
    for(int i=n*m;i>=1;i--)inv[i-1]=1ll*inv[i]*i%mod;
    f[0][0]=1;
    for(int i=1;i<=n;i++)
        for(int j=0;j<=i;j++)
        {
            f[i][j]=f[i-1][j];
            if(!j)continue;
            Mod(f[i][j],1ll*f[i][j-1]*(n-j+1)%mod*C(n-i+(n-j+1)*(m-1)-1,m-2)%mod);
        }
    printf("%d",f[n][n]);
    return 0;
}