BZOJ 5123([Lydsy12月赛]线段树的匹配-记忆化搜索)

题意：给定一棵表示 [1, n] 的线段树，请求出它的最大匹配中有多少条边，并求出有多少种最大匹配的方案。

显然线段树形态由长度唯一确定，因此可以用（长度，根节点是否和子树的节点匹配）作为状态进行记忆化搜索

#include<bits/stdc++.h> 
using namespace std;
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#define
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma
#define
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
typedef pair<ll,ll> pll;
map<pair<ll,bool>  ,pll > f;
void upd(pair<ll,ll> &a, pair<ll,ll> b) {
    if (a.fi!=b.fi) a=max(a,b);
    else a.se=(a.se+b.se)%F;
}
pll composed(pll a,pll b) {
    return mp(a.fi+b.fi, (a.se*b.se)%F);
}
pll Max(pll a,pll b) {
    if (a.fi!=b.fi) return max(a,b);
    else return mp(a.fi,(a.se+b.se)%F);
}
pll dfs(ll x,bool b) {
    if(f.find(mp(x,b))!=f.end()) return f[mp(x,b)];
    pll l0=dfs(x>>1,0),l1=dfs(x>>1,1),r0=dfs((x-(x>>1)),0),r1=dfs((x-(x>>1)),1);
    pll c00=composed(l0,r0),c01=composed(l0,r1),c10=composed(l1,r0),c11=composed(l1,r1);
    if (!b) {
        f[mp(x,0)] = Max(c00,Max(c01,Max(c10,c11)));

    }else {
        f[mp(x,1)] = Max(composed(c00 , mp(1,2)) ,composed(Max(c01,c10), mp(1,1)) );
    }
    return f[mp(x,b)];
}
int main()
{
//  freopen("bzoj5123.in","r",stdin);
//  freopen(".out","w",stdout);
    f[mp(1,0)]=mp(0,1);
    f[mp(1,1)]=mp(-1000,0);
    ll n;
    cin>>n;
    pll ans=dfs(n,0),ans2=dfs(n,1);

    upd(ans,ans2);
    cout<<ans.fi<<' '<<ans.se<<endl;
    return 0;
}