answer
Tree dp, design condition is as follows:
设 DP[u][i][0] 表示 以点 u 为根的子树 最大匹配数模 m 为 i 时,且 u 点没有匹配的方案数
DP[u][i][1] 表示 以点 u 为根的子树 最大匹配数模 m 为 i 时,且 u 点匹配上的方案数
Recursive formula is as follows:
DP[u][k][0](不匹配该节点) += ∑ [i+j==k] 2 * DP[u][i][0] * DP[v][j][1](此时u->这条边连不连都无所谓,所以*2) + DP[u][i][0] * DP[v][j][0](儿子已近匹配了)
DP[u][k][1](该节点已经连了边) += ∑ [i+j==k] 2 * DP[u][i][1] * ( DP[v][j][0] + DP[v][j][1] )
DP[u][k][1] (该节点现在正要连边)+= ∑ [i+j==k-1(预留出一个位用于匹配)] DP[u][i][0] * DP[v][j][0](此时这条边必须存在并且u,v都不能匹配别的边)
Reference from
https://www.cnblogs.com/nicetomeetu/p/7360556.html
Code
#include <the iostream> #include <CString> #include <cstdio> #include <Vector> the using namespace STD; #define int Long Long #define MOD 998 244 353 #define N 50001 #define M 410 Vector < int > VEC [N]; int DP [N] [M] [ 2 ], size [N] / * record nodes to up to match the number edge * /, TEMP [M] [ 2 ] / * DP array in the middle of an update can not be changed, so with this array Instead of * /, m; void the Add ( int U, int V) // update the array using edge { Memset (TEMP, 0 , the sizeof(TEMP)); for ( int I = 0 ; I <= size [U]; I ++ ) // where the time complexity can prove to n-* m { for ( int J = 0 ; J <= size [V]; J ++ ) { TEMP [I + J] [ 0 ] + = 2 * DP [U] [I] [ 0 ] * DP [V] [J] [ . 1 ] + DP [U] [I] [ 0 ] * DP [V] [J] [ 0 ]; TEMP [I + J] [ 0 ]% = MOD; TEMP [I + J] [ . 1 ] + = 2 * DP [U] [I] [ . 1 ] * (DP [ V] [J] [ 0]+dp[v][j][1]); temp[i+j][1]%=mod; temp[i+j+1][1]+=dp[u][i][0]*dp[v][j][0]; temp[i+j+1][1]%=mod; } } for(int i=0;i<m;i++)//将temp复制到dp内 { dp[u][i][0]=(temp[i][0]+temp[i+m][0])%mod;//i+j有可能超过n dp[u][i][1]=(temp[i][1]+temp[i+m][1])%mod; } size[u]=min(size[u]+size[v],m);//size[u]不能超过m,否则会数组出界。 } void dfs(int id,int from) { size[id]=1; dp[id][0][0]=1; for(int i=0;i<vec[id].size();i++) { int to=vec[id][i]; if(to==from) continue; dfs(to,id);//树形dp的惯例,自底向上更新 add(id,to); } } signed main() { int n; cin>>n>>m; for(int i=1;i<n;i++) { int a,b; scanf("%lld%lld",&a,&b); vec[a].push_back(b); vec[b].push_back(a); } dfs(1,0); cout<<(dp[1][0][0]+dp[1][0][1])%mod; }