題目鏈接

http://codeforces.com/contest/1276/problem/D

題解

我什么DP都不會做，吃棗藥丸……
設\(f_{u,j}\)表示\(u\)子樹內，\(j=0\)要求\(u\)點在輪到其父邊之前被刪，\(j=1\)要求\(u\)點被其父邊刪掉，\(j=2\)要求\(u\)點在其父邊之后被刪或者最后沒有被刪。
轉移: 設兒子有\(s\)個，分別為\(v_1,v_2,...,v_s\), 且按邊的編號從小到大排序，父邊編號位于\(d\)和\((d+1)\)之間。
枚舉被哪條邊刪除。
\[f_{u,0}=\sum^d_{i=1}(\prod^{i-1}_{j=1}(f_{v_j,0}+f_{v_j,1})\cdot f_{v_i,2}\cdot \prod^s_{j=i+1}(f_{v_j,0}+f_{v_j,2}))\]
\[f_{u,1}=\prod^d_{i=1}(f_{v_j,0}+f_{v_j,1})\cdot \prod^s_{i=d+1}(f_{v_j,0}+f_{v_j,2})\]
\[f_{u,2}=\sum^s_{i=d+1}(\prod^{i-1}_{j=1}(f_{v_j,0}+f_{v_j,1})\cdot f_{v_i,2}\cdot \prod^s_{j=i+1}(f_{v_j,0}+f_{v_j,2}))+\prod^{s}_{i=1}(f_{v_j,0}+f_{v_j,1})\]
維護前后綴積即可。
時間復雜度\(O(n)\).

代碼

#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; }

總結

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