2021 ICPC regional competition (Shanghai) g.edge groups (tree DP)

Edge set

quote lwz_159 My blog

$Code:$

#include<bits/stdc++.h>
#include<unordered_map>
#define mem(a,b) memset(a,b,sizeof a)
#define cinios (ios::sync_with_stdio(false),cin.tie(0),cout.tie(0))
#define sca scanf
#define pri printf
#define forr(a,b,c) for(int a=b;a<=c;a++)
#define rfor(a,b,c) for(int a=b;a>=c;a--)
#define endl "\n"
//[ Blog address ]：https://blog.csdn.net/weixin_51797626?t=1
using namespace std;
inline void read(int& x) {
     x = 0; int f = 1; char ch = getchar(); while (ch < '0' || ch > '9') {
     if (ch == '-') f = -1; ch = getchar(); } while (ch >= '0' && ch <= '9') {
     x = x * 10 + (ch - '0'); ch = getchar(); } x *= f; }
void write(int x) {
     if (x < 0) putchar('-'), x = -x; if (x >= 10) write(x / 10); putchar(x % 10 + '0'); }

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;

const int N = 100010, M = 200010, MM = N;
int INF = 0x3f3f3f3f, mod = 998244353;
ll LNF = 0x3f3f3f3f3f3f3f3f;
int n, m, k, T, S, D;
int h[N], ne[M], e[M], idx;
ll dp[N];

inline void add(int a, int b) {
    
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

// Determine the edges that can contribute according to the number of edges of the subtree where the child node is located 
// If the number of sides is even （ Pairing within oneself ）, The edges of the current node and child nodes contribute 
// Otherwise, odd numbers will occupy this side 

bool dfs(int x, int fa) {
    
    int cnt = 0;
    dp[x] = 1;

    for (int i = h[x]; ~i; i = ne[i]) {
    
        int j = e[i];
        if (j == fa)continue;
        if (!dfs(j, x))cnt++;
        dp[x] = (dp[x] * dp[j]) % mod;
    }
    for (int i = 1; i <= cnt; i += 2)dp[x] = (dp[x] * i) % mod;
    return cnt & 1;
}

int main() {
    
    cinios;

    cin >> n;
    mem(h, -1);

    forr(i, 1, n - 1) {
    
        int a, b;
        cin >> a >> b;
        add(a, b), add(b, a);
    }
    dfs(1, -1);

    cout << dp[1];

    return 0;
}
/* */