UOJ 185【ZJOI2016】小星星

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求点数相同的一个无向图与一棵树有多少种不同的映射，使得树的边集为图的边集的子集。

首先orz吨爷

可以说是因为吨爷才去学的子集卷积(vfleaking)==结果过不掉
这样写树形dp感觉还行吧（燃烧的内存），好像这几次利用单调性什么的，最后都没用到+卷积，只是用or卷积，以后会遇上吗？

O(2n∗n3)纯净TLE版

#include <set>#include <ctime>#include <queue>#include <cstdio>#include <bitset>#include <cctype>#include <bitset>#include <cstdlib>#include <cassert>#include <cstring>#include <iostream>#include <algorithm>#define inf (1<<30)#define INF (1ll<<62)#define fi first#define se second#define rep(x,s,t) for(register int x=s,t_=t;x<t_;++x)#define per(x,s,t) for(register int x=t-1,s_=s;x>=s_;--x)#define prt(x) cout<<#x<<":"<<x<<" "#define prtn(x) cout<<#x<<":"<<x<<endl#define travel(x) for(int I=last[x],to;to=e[I].to,I;I=e[I].nxt)#define pb(x) push_back(x)#define hash asfmaljkg#define rank asfjhgskjf#define y1 asggnja#define y2 slfvmusing namespace std;typedef long long ll;typedef pair<int,int> ii;template<class T>void sc(T &x){    int f=1;char c;x=0;    while(c=getchar(),c<48)if(c=='-')f=-1;    do x=x*10+(c^48);    while(c=getchar(),c>47);    x*=f;}template<class T>void nt(T x){    if(!x)return;    nt(x/10);    putchar(x%10+'0');}template<class T>void pt(T x){    if(x<0)putchar('-'),x=-x;    if(!x)putchar('0');    else nt(x);}template<class T>void ptn(T x){    pt(x);putchar('\n');}template<class T>void pts(T x){    pt(x);putchar(' ');}template<class T>inline void Max(T &x,T y){if(x<y)x=y;}template<class T>inline void Min(T &x,T y){if(x>y)x=y;}const int maxn=18;int n,m;int b[maxn][maxn];int last[maxn],ecnt;struct Edge{    int to,nxt;}e[maxn<<1];void ins(int u,int v){    e[++ecnt]=(Edge){v,last[u]};    last[u]=ecnt;}ll dp[maxn][maxn][1<<maxn];ll dp2[1<<maxn];void ormul(ll *a,ll *b,ll *c){    static ll A[1<<maxn],B[1<<maxn];    rep(i,0,1<<n)A[i]=a[i],B[i]=b[i];    rep(i,0,n)rep(j,1,1<<n)        if(j>>i&1){            A[j]+=A[j^1<<i];            B[j]+=B[j^1<<i];        }    rep(i,0,1<<n)c[i]=A[i]*B[i];    rep(i,0,n)rep(j,1,1<<n)        if(j>>i&1)c[j]-=c[j^1<<i];}int cnt=0;void dfs(int x,int f){    travel(x)if(to!=f)dfs(to,x);    rep(i,0,n){        dp[x][i][1<<i]=1;        travel(x)if(to!=f){            rep(k,0,1<<n)dp2[k]=0;            rep(j,0,n)if(b[i][j])                rep(k,0,1<<n)dp2[k]+=dp[to][j][k];            ormul(dp[x][i],dp2,dp[x][i]);        }    }}int main(){//  freopen("pro.in","r",stdin);//  freopen("ex_star2.in","r",stdin);//  freopen("chk.out","w",stdout);    sc(n);sc(m);    rep(i,0,m){        int u,v;        sc(u);sc(v);--u;--v;        b[u][v]=b[v][u]=true;    }    rep(i,1,n){        int u,v;        sc(u);sc(v);--u;--v;        ins(u,v);ins(v,u);    }    dfs(0,-1);    ll ans=0;int all=(1<<n)-1;    rep(i,0,n)ans+=dp[0][i][all];    ptn(ans);    return 0;}

看了StilWell的代码之后发现：
小范围直接卷会快很多(甚至都可以直接AC了）
不仅仅只有vfk版卷积

有一个更通用的构造，FWT(Fast Walsh-Hadamard Transform)(zhuohan123,Picks)

$t f x o r (A) = (t f x o r (A 0 + A 1), t f x o r (A 0 - A 1)) u t f x o r (A) = (u t f x o r (A 0 + A 1 2), u t f x o r (A 0 - A 1 2))$
$t f a n d (A) = (t f a n d (A 0 + A 1), t f a n d (A 1)) u t f a n d (A) = (u t f a n d (A 0 - A 1), u t f a n d (A 1))$
$t f o r (A) = (t f o r (A 0), t f o r (A 1 + A 0)) u t f o r (A) = (u t f o r (A 0), u t f o r (A 1 - A 0))$

笔记：
考虑一位时，

x o r C 0 = ( A 0 + A 1 ) \times ( B 0 + B 1 ) + ( A 0 - A 1 ) \times ( B 0 - B 1 ) 2 = A 0 \times B 0 + A 1 \times B 1 x o r C 1 = ( A 0 + A 1 ) \times ( B 0 + B 1 ) - ( A 0 - A 1 ) \times ( B 0 - B 1 ) 2 = A 0 \times B 1 + A 1 \times B 0

a n d C 0 = (A 0 + A 1) \times (B 0 + B 1) - A 1 \times B 1 = \dots a n d C 1 = A 1 \times B 1

o r C 0 = A 0 \times B 0 o r C 1 = (A 0 + A 1) \times (B 0 + B 1) - A 0 \times B 0 = \dots

就是容斥==
然后很多位的的时候再来个大容斥
这就是FWT，用来解决位运算的卷积的构造

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