The 15th Chinese Northeast Collegiate C. Vertex Deletion(树形dp)

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对于节点 u u u来说(分别对应状态 0 , 1 , 2 0,1,2 0,1,2)

定义 f [ i ] [ 0 / 1 / 2 ] f[i][0/1/2] f[i][0/1/2]表示当 i i i分别满足以下三个条件时,子树内节点全部符合要求时的答案

Ⅰ.如果删去这个节点,那么 u u u一定满足条件

Ⅱ.如果不删去 u u u,且所有儿子被删掉.那么此时 u u u还没满足条件,需要父亲也不被删去

Ⅲ.如果不删去 u u u,且所有儿子中至少有一个没删掉.那么 u u u满足条件,父亲怎样无所谓

可以发现这个设计非常合理,比如

f [ u ] [ 0 ] = ∏ v ∈ s o n u f [ v ] [ 0 ] + f [ v ] [ 2 ] f[u][0]=\prod\limits_{v\in son_u}f[v][0]+f[v][2] f[u][0]=v∈sonu∏f[v][0]+f[v][2]

f [ u ] [ 1 ] = ∏ v ∈ s o n u f [ v ] [ 0 ] f[u][1]=\prod\limits_{v\in son_u}f[v][0] f[u][1]=v∈sonu∏f[v][0]

但是发现转移 f [ u ] [ 2 ] f[u][2] f[u][2]时有点变化,我们可以采用容斥的办法

也就是所有方案数-不合法的方案数,即为

f [ u ] [ 2 ] = ∏ v ∈ s o n u ( f [ v ] [ 0 ] + f [ v ] [ 1 ] + f [ v ] [ 2 ] ) − ∏ v ∈ s o n u f [ v ] [ 0 ] f[u][2]=\prod\limits_{v\in son_u}(f[v][0]+f[v][1]+f[v][2])-\prod\limits_{v\in son_u}f[v][0] f[u][2]=v∈sonu∏(f[v][0]+f[v][1]+f[v][2])−v∈sonu∏f[v][0]

显然 f [ r o o t ] [ 0 ] + f [ r o o t ] [ 2 ] f[root][0]+f[root][2] f[root][0]+f[root][2]即是答案

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int mod = 998244353;
const int maxn = 1e5+10;
int n,f[maxn][3];
vector<int>vec[maxn];
/*
f[i][0]表示自己被删掉
f[i][1]表示自己没删去,且没有儿子和自己相连
f[i][2]表示自己没删去,且至少有一个儿子和自己相连
*/
void dfs(int u,int fa)
{f[u][0] = f[u][1] = f[u][2] = 1;for(auto v:vec[u] ){if( v==fa )   continue;dfs( v,u );f[u][0] = ( f[v][0]+f[v][2] )%mod*f[u][0]%mod;f[u][1] = f[u][1]*f[v][0]%mod;f[u][2] = ( f[v][0]+f[v][1]+f[v][2] )%mod*f[u][2]%mod; }f[u][2] = ( f[u][2]-f[u][1] )%mod;//f[u][2]为总方案数-所有儿子都被删去的方案数
}
signed main()
{int t; cin >> t;while( t-- ){cin >> n;for(int i=1;i<n;i++){int l,r; scanf("%lld%lld",&l,&r);vec[l].push_back( r ), vec[r].push_back( l );  }   dfs( 1,0 );cout << (( f[1][0]+f[1][2] )%mod+mod)%mod << endl;for(int i=1;i<=n;i++) vec[i].clear();}
}

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