Atcoder ARC101 E 樹dp

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https://arc101.contest.atcoder.jp/tasks/arc101_c

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題解是也是dp，好像是容斥做的，但是看不懂，而且也好像沒講怎么變n^2，看了寫大佬的代碼，自己理解了一下

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
#define X           first
#define Y           second
#define pb          push_back
#define mp          make_pair
#define SZ(X)       (X.size())
#define mst(a,b)    memset((a),(b),sizeof(a))
#define lowbit(a)   ((a)&(-a))
using namespace std;
typedef unsigned long long ull;
typedef long long LL;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
const int mod=1e9+7;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int maxn=5500;
inline int add(int x,int y){if((x+=y)>=mod)x-=mod;return x;
}
inline int mul(int x,int y){return (ll)x*y%mod;
}
inline int sub(int x,int y){if((x-=y)<0)x+=mod;return x;
}
int ci[maxn];
vector<int>to[maxn];
int dp[maxn][maxn];//dp[pos][i] 子樹,有i個點未匹配的合法方案//除了根之外的子樹自匹配完是不合法的,所以-dp[pos][0]表示以pos為根的子樹匹配完,但pos之下子樹各自之間未匹配完
int sz[maxn],uu[maxn];
void dfs(int pos,int fa){sz[pos]=1;dp[pos][1]=1;for(int d:to[pos])if(d!=fa){dfs(d,pos);for(int i=0;i<=sz[pos]+sz[d];++i)uu[i]=0;for(int i=0;i<=sz[pos];++i)for(int j=0;j<=sz[d];++j)uu[i+j]=add(uu[i+j],mul(dp[pos][i],dp[d][j]));sz[pos]+=sz[d];for(int i=0;i<=sz[pos];++i)dp[pos][i]=uu[i];}for(int i=2;i<=sz[pos];i+=2)dp[pos][0]=sub(dp[pos][0],mul(dp[pos][i],ci[i]));
}
int main() {
#ifdef localfreopen("in.txt", "r", stdin);
#endif // localios::sync_with_stdio(0);cin.tie(0);cout.tie(0);ci[0]=1;for(int i=2;i<maxn;i+=2)ci[i]=mul(ci[i-2],i-1);int n;cin>>n;for(int i=1;i<n;++i){int a,b;cin>>a>>b;to[a].push_back(b);to[b].push_back(a);}dfs(1,0);cout<<(mod-dp[1][0]);return 0;
}

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