五校联考模拟赛Day2T2矩阵(容斥原理)

题意

$n * m$的网格，对其进行黑白染色，问每一行每一列至少有一个黑格子的方案数。

Sol

考场上只会$n^3$的dp，还和指数级枚举一个分qwq

设$f[i][j]$表示到了第$i$行，已经有$j$列被染黑，然后暴力转移上一行有几个黑格子

正解是容斥

首先固定好列，也就是保证每一列都有一个黑格子

这样的方案是$(2^N - 1) ^M$

然后容斥行

组合数暴力算即可

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
#include<iostream>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long 
#define LL long long 
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9 + 10, mod = 1e9 + 7;
const double eps = 1e-9;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    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;
}
int N, M;
LL fac[MAXN], ifac[MAXN], po2[MAXN];
LL fastpow(int a, int p) {
    LL base = 1;
    while(p) {
        if(p & 1) base = (base * a) % mod;
        a = (a * a) % mod; p >>= 1;
    }
    return base % mod;
}
LL C(int N, int M) {
    return fac[N] * ifac[M] % mod * ifac[N - M] % mod;
}
main() {
    N = 2 * 1e5;
    fac[0] = 1; po2[0] = 1;
    for(int i = 1; i <= N; i++) fac[i] = i * fac[i - 1] % mod, po2[i] = (po2[i - 1] * 2) % mod;
    ifac[N] = fastpow(fac[N], mod - 2); 
    for(int i = N; i >= 1; i--) 
        ifac[i - 1] = (ifac[i] % mod * i) % mod;
    N = read(); M = read();
    int d = 1; LL ans = 0;
    for(int i = 0; i <= N; i++, d *= -1) 
        ans = (ans + d * C(N, i) * fastpow((po2[N - i] - 1 + mod) % mod, M) % mod + mod) % mod;
    cout << ans;
    return 0;
}