LOJ2540 随机算法

题目链接
看到\(n\leq 20\),马上想到状压\(dp\).
考虑用\(f[S][i]\)表示集合\(S\)已经被考虑过了,独立集大小为\(i\)的方案数.
显然,这个集合\(S\)的最外层显然都没有被选.
考虑如何转移.
枚举一个\(j\notin S\),那么独立集大小显然\(+1\),然后所有和\(j\)相连的点都不能选了.
那么用\(w[j]\)记录与\(j\)相邻的集合,然后就可以转移了.
\(f[S\cup j][i+1]\leftarrow f[S\cup j][i+1]+f[S][i]*A_{n-|S|-1}^{|w[j]|-|w[j]\cap S|-1}\)
时间复杂度看似是\(O(2^n*n^2)\)的,但是因为跑不满,所以跑得过
代码如下
其实也没有什么好贴的了

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#define N (22)
#define P (998244353)
#define inf (0x7f7f7f7f)
#define rg register int
#define Label puts("NAIVE")
#define spa print(' ')
#define ent print('\n')
#define rand() (((rand())<<(15))^(rand()))
typedef long double ld;
typedef long long LL;
typedef unsigned long long ull;
using namespace std;
inline char read(){
    static const int IN_LEN=1000000;
    static char buf[IN_LEN],*s,*t;
    return (s==t?t=(s=buf)+fread(buf,1,IN_LEN,stdin),(s==t?-1:*s++):*s++);
}
template<class T>
inline void read(T &x){
    static bool iosig;
    static char c;
    for(iosig=false,c=read();!isdigit(c);c=read()){
        if(c=='-')iosig=true;
        if(c==-1)return;
    }
    for(x=0;isdigit(c);c=read())x=((x+(x<<2))<<1)+(c^'0');
    if(iosig)x=-x;
}
inline char readchar(){
    static char c;
    for(c=read();!isalpha(c);c=read())
    if(c==-1)return 0;
    return c;
}
const int OUT_LEN = 10000000;
char obuf[OUT_LEN],*ooh=obuf;
inline void print(char c) {
    if(ooh==obuf+OUT_LEN)fwrite(obuf,1,OUT_LEN,stdout),ooh=obuf;
    *ooh++=c;
}
template<class T>
inline void print(T x){
    static int buf[30],cnt;
    if(x==0)print('0');
    else{
        if(x<0)print('-'),x=-x;
        for(cnt=0;x;x/=10)buf[++cnt]=x%10+48;
        while(cnt)print((char)buf[cnt--]);
    }
}
inline void flush(){fwrite(obuf,1,ooh-obuf,stdout);}
int n,s[N][N],w[N],Lim,cnt[(1<<N)],m;
LL f[1<<N][N+1],jc[N],inv[N];
int popcount(int x){
    int cnt=0;
    for(;x;x&=(x-1))cnt++;
    return cnt;
}
LL ksm(LL a,int p){
    LL res=1;
    while(p){
        if(p&1)res=(res*a)%P;
        a=(a*a)%P,p>>=1;
    }
    return res;
}
LL A(int n,int m){
    return jc[n]*inv[n-m]%P;
}
int main(){
    read(n),read(m);
    Lim=(1<<n)-1,jc[0]=inv[0]=1;
    for(int i=1;i<=m;i++){
        int x,y;
        read(x),read(y);
        s[x][y]=s[y][x]=1;
    }
    for(int i=1;i<=n;i++)
    jc[i]=(jc[i-1]*i)%P,inv[i]=ksm(jc[i],P-2)%P;
    for(int i=1;i<=n;i++)w[i]=(1<<i-1);
    for(int i=1;i<=n;i++)
    for(int j=i+1;j<=n;j++)
    if(s[i][j])w[i]+=(1<<j-1),w[j]+=(1<<i-1);
    f[0][0]=1;
    for(int i=0;i<=Lim;i++)
    cnt[i]=popcount(i);
    for(int i=0;i<=Lim;i++)
    for(int k=1;k<=n;k++)
    if(((1<<k-1)&i)==0){
        int v=i|w[k];
        int x=cnt[w[k]]-cnt[i&w[k]]-1;
        int t=n-cnt[i]-1;
        for(int j=0;j<=n;j++)
        if(f[i][j])f[v][j+1]=(f[v][j+1]+f[i][j]*A(t,x)%P)%P;
    }
    for(int i=n;i>=0;i--)
    if(f[Lim][i]){
        printf("%lld\n",(f[Lim][i]*inv[n])%P);
        return 0;
    }
}

转载于:https://www.cnblogs.com/Romeolong/p/10058991.html