Codeforces

Luogu

分析

显然可以考虑容斥。枚举 $i$ 行 $j$ 列的最小值不为 $1$，可以得到答案为

$$ \sum_{i=0}^n\sum_{j=0}^n(-1)^{i+j}{n\choose i}{n\choose j}(k-1)^{ni+nj-ij}k^{n^2-ni-nj+ij} $$

代码

// ===================================
//   author: M_sea
//   website: http://m-sea-blog.com/
// ===================================
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#define re register
using namespace std;

inline int read() {
    int X=0,w=1; char c=getchar();
    while (c<'0'||c>'9') { if (c=='-') w=-1; c=getchar(); }
    while (c>='0'&&c<='9') X=X*10+c-'0',c=getchar();
    return X*w;
}

const int mod=1e9+7;
inline int qpow(int a,int b) { int c=1;
    for (;b;b>>=1,a=1ll*a*a%mod) if (b&1) c=1ll*c*a%mod;
    return c;
}

const int N=250+10;

int fac[N],ifac[N];
inline int C(int n,int m) {
    return 1ll*fac[n]*ifac[m]%mod*ifac[n-m]%mod;
}

int main() {
    int n=read(),k=read(),ans=0;
    fac[0]=1;
    for (re int i=1;i<=n;++i) fac[i]=1ll*fac[i-1]*i%mod;
    ifac[n]=qpow(fac[n],mod-2);
    for (re int i=n;i;--i) ifac[i-1]=1ll*ifac[i]*i%mod;
    for (re int i=0;i<=n;++i)
        for (re int j=0;j<=n;++j) {
            int b=n*i+n*j-i*j;
            int w=1ll*C(n,i)*C(n,j)%mod*qpow(k-1,b)%mod*qpow(k,n*n-b)%mod;
            if ((i+j)&1) ans=(ans-w+mod)%mod;
            else ans=(ans+w)%mod;
        }
    printf("%d\n",ans);
    return 0;
}