loj2540 「PKUWC2018」随机算法 【状压dp】

题目链接

loj2540

题解

有一个朴素三进制状压\(dp\)，考虑当前点三种状态：没考虑过，被选入集合，被排除

就有了\(O(n3^{n})\)的转移

但这样不优，我们考虑优化状态

设\(f[i][S]\)表示独立集大小为\(i\)，不可选集合为\(S\)【要么是已经在独立集中，要么已经被排除了】

那么剩余点都是可选的

就枚举剩余点\(u\)，记\(u\)相邻的集合为\(S_u\)，那么当\(u\)加入后，集合\(S_u\)的点都不能选，但是由于所有点都会加入排列之中，\(S_u\)中除了\(S\)中的点，剩余点的点会排列在\(u\)之后，那么有

\[f[i + 1][S \cup S_u \cup \{u\}] += A_{n - |S| - 1}^{|S_u| - |S_u \cap S|}f[i][S]
\]

复杂度\(O(n^{2}2^{n})\)

但是有很多没用的状态，所以可以过

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 22,maxm = (1 << 21),INF = 1000000000,P = 998244353;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
int n,m,G[maxn],f[maxn][maxm],fac[maxn],fv[maxn];
inline int qpow(int a,int b){
	int re = 1;
	for (; b; b >>= 1,a = 1ll * a * a % P)
		if (b & 1) re = 1ll * re * a % P;
	return re;
}
inline int cal(int s){
	int re = 0;
	while (s) re += (s & 1),s >>= 1;
	return re;
}
inline int A(int n,int m){
	return 1ll * fac[n] * fv[n - m] % P;
}
int main(){
	fac[0] = 1; for (int i = 1; i <= 20; i++) fac[i] = 1ll * fac[i - 1] * i % P;
	fv[20] = qpow(fac[20],P - 2); fv[0] = 1;
	for (int i = 19; i; i--) fv[i] = 1ll * fv[i + 1] * (i + 1) % P;
	n = read(); m = read(); int u,v;
	while (m--){
		u = read(),v = read();
		G[u] |= (1 << v - 1);
		G[v] |= (1 << u - 1);
	}
	int maxv = (1 << n) - 1;
	f[0][0] = 1;
	for (int i = 0; i < n; i++){
		for (int s = 0; s <= maxv; s++){
			if (!f[i][s]) continue;
			int tot = cal(s);
			for (int u = 1; u <= n; u++)
				if (!(s & (1 << u - 1))){
					int siz = cal(G[u]) - cal(s & G[u]),e = s | G[u] | (1 << u - 1);
					f[i + 1][e] = (f[i + 1][e] + 1ll * f[i][s] * A(n - tot - 1,siz) % P) % P;
				}
		}
	}
	for (int i = n; i; i--)
		if (f[i][maxv]){
			printf("%lld\n",1ll * f[i][maxv] * fv[n] % P);
			break;
		}
	return 0;
}