BZOJ5123 线段树的匹配（树形dp）

　　线段树的任意一棵子树都相当于节点数与该子树相同的线段树。于是假装在树形dp即可，记忆化搜索实现，有效状态数是logn级别的。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
#define ll long long
#define P 998244353
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
ll n;
map<ll,ll> f[],g[];
void solve(ll n)
{
    if (g[].find(n)!=g[].end()) return;
    ll lson=n+>>,rson=n-lson;
    solve(lson),solve(rson);
    f[][n]=max(f[][lson],f[][lson])+max(f[][rson],f[][rson]);
    f[][n]=max(f[][lson]+max(f[][rson],f[][rson]),f[][rson]+max(f[][lson],f[][lson]))+;
    ll x=f[][lson]==f[][lson]?g[][lson]+g[][lson]:(f[][lson]>f[][lson]?g[][lson]:g[][lson]);
    ll y=f[][rson]==f[][rson]?g[][rson]+g[][rson]:(f[][rson]>f[][rson]?g[][rson]:g[][rson]);
    g[][n]=x*y%P;
    if (f[][lson]+max(f[][rson],f[][rson])==f[][rson]+max(f[][lson],f[][lson]))
    g[][n]=(g[][lson]*y+x*g[][rson])%P;
    else if (f[][lson]+max(f[][rson],f[][rson])>f[][rson]+max(f[][lson],f[][lson])) g[][n]=g[][lson]*y%P;
    else g[][n]=x*g[][rson]%P;
    return;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj5123.in","r",stdin);
    freopen("bzoj5123.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    cin>>n;f[][]=,f[][]=-n,g[][]=,g[][]=;
    solve(n);
    if (f[][n]==f[][n]) cout<<f[][n]<<' '<<(g[][n]+g[][n])%P;
    else if (f[][n]>f[][n]) cout<<f[][n]<<' '<<g[][n];
    else cout<<f[][n]<<' '<<g[][n];
    return ;
}