题目链接
http://codeforces.com/contest/1276/problem/D
题解
我什么DP都不会做，吃枣药丸……
设\(f_{u,j}\)表示\(u\)子树内，\(j=0\)要求\(u\)点在轮到其父边之前被删，\(j=1\)要求\(u\)点被其父边删掉，\(j=2\)要求\(u\)点在其父边之后被删或者最后没有被删。
转移: 设儿子有\(s\)个，分别为\(v_1,v_2,...,v_s\), 且按边的编号从小到大排序，父边编号位于\(d\)和\((d+1)\)之间。
枚举被哪条边删除。
\[f_{u,0}=\sum^d_{i=1}(\prod^{i-1}_{j=1}(f_{v_j,0}+f_{v_j,1})\cdot f_{v_i,2}\cdot \prod^s_{j=i+1}(f_{v_j,0}+f_{v_j,2}))
\]
\[f_{u,1}=\prod^d_{i=1}(f_{v_j,0}+f_{v_j,1})\cdot \prod^s_{i=d+1}(f_{v_j,0}+f_{v_j,2})
\]
\[f_{u,2}=\sum^s_{i=d+1}(\prod^{i-1}_{j=1}(f_{v_j,0}+f_{v_j,1})\cdot f_{v_i,2}\cdot \prod^s_{j=i+1}(f_{v_j,0}+f_{v_j,2}))+\prod^{s}_{i=1}(f_{v_j,0}+f_{v_j,1})
\]
维护前后缀积即可。
时间复杂度\(O(n)\).
代码
#include<bits/stdc++.h>
#define llong long long
#define pii pair<int,int>
#define mkpr make_pair
using namespace std;
inline int read()
{
int x = 0,f = 1; char ch = getchar();
for(;!isdigit(ch);ch=getchar()) {if(ch=='-') f = -1;}
for(; isdigit(ch);ch=getchar()) {x = x*10+ch-48;}
return x*f;
}
const int N = 2e5;
const int P = 998244353;
vector<pii> adj[N+3];
int fa[N+3],fae[N+3];
llong aux1[N+3],aux2[N+3];
llong f[N+3][3];
int n,en;
void dfs(int u)
{
sort(adj[u].begin(),adj[u].end()); int faid = -1,adjn = adj[u].size();
for(int i=0; i<adj[u].size(); i++)
{
int o = adj[u][i].first,v = adj[u][i].second;
if(v==fa[u]) {faid = i; continue;} fa[v] = u,fae[v] = o;
dfs(v);
}
aux1[0] = 1ll;
for(int i=0; i<adj[u].size(); i++)
{
int v = adj[u][i].second; if(v==fa[u]) {aux1[i+1] = aux1[i]; continue;}
aux1[i+1] = aux1[i]*(f[v][0]+f[v][1])%P;
}
aux2[adj[u].size()+1] = 1ll;
for(int i=(int)adj[u].size()-1; i>=0; i--)
{
int v = adj[u][i].second; if(v==fa[u]) {aux2[i+1] = aux2[i+2]; continue;}
aux2[i+1] = aux2[i+2]*(f[v][0]+f[v][2])%P;
}
f[u][0] = 0ll;
for(int i=0; i<faid; i++)
{
int v = adj[u][i].second;
llong tmp = aux1[i]*f[v][2]%P*aux2[i+2]%P; f[u][0] = (f[u][0]+tmp)%P;
}
if(faid!=-1) {f[u][1] = aux1[faid]*aux2[faid+2]%P;}
f[u][2] = 0ll;
for(int i=faid+1; i<adj[u].size(); i++)
{
int v = adj[u][i].second;
llong tmp = aux1[i]*f[v][2]%P*aux2[i+2]%P; f[u][2] = (f[u][2]+tmp)%P;
}
f[u][2] = (f[u][2]+aux1[adjn])%P;
}
int main()
{
scanf("%d",&n);
for(int i=1; i<n; i++)
{
int u,v; scanf("%d%d",&u,&v);
adj[u].push_back(mkpr(i,v)); adj[v].push_back(mkpr(i,u));
}
dfs(1);
printf("%I64d\n",f[1][2]);
return 0;
}
