GP of China H Inner Product: 边分治 + 虚树dp
题意
给出两棵树 T T T和 T ′ T' T′，求
∑ i , j ∈ [ 1 , n ] d i s ( i , j ) ∗ d i s ′ ( i , j ) \sum_{i,j \in [1,n]}{dis(i,j) * dis'(i,j)} i,j∈[1,n]∑dis(i,j)∗dis′(i,j)
题解
对 T T T进行边分治，当前分治的边为 < u , v >
A N S = ∑ i ∈ L , j ∈ R ( d i s ( i , u ) + d i s ( j , v ) + w ) ⋅ d i s ′ ( i , j ) = ∑ i ∈ L , j ∈ R ( d i s [ i ] + d i s [ j ] ) ⋅ d i s ′ ( i , j ) + w ⋅ ∑ i ∈ L , j ∈ R d i s ′ ( i , j ) = ∑ i ∈ L , j ∈ R ( d i s [ i ] + d i s [ j ] ) ⋅ ( d e p ′ [ i ] + d e p ′ [ j ] − 2 ⋅ d e p ′ [ l c a ′ ( i , j ) ] ) + w ⋅ ∑ i ∈ L , j ∈ R d i s ′ ( i , j ) = ∑ i ∈ L , j ∈ R ( d i s [ i ] + d i s [ j ] ) ⋅ ( d e p ′ [ i ] + d e p ′ [ j ] ) − 2 ⋅ ∑ i ∈ L , j ∈ R ( d i s [ i ] + d i s [ j ] ) ⋅ d e p ′ [ l c a ′ ( i , j ) ] + w ⋅ ∑ i ∈ L , j ∈ R d i s ′ ( i , j ) = ∑ i ∈ L , j ∈ R ( d i s [ i ] ⋅ d e p ′ [ i ] + d i s [ j ] ⋅ d e p ′ [ j ] ) + ∑ i ∈ L , j ∈ R ( d i s [ i ] ⋅ d e p ′ [ j ] + d i s [ j ] ⋅ d e p ′ [ i ] ) − 2 ⋅ ∑ i ∈ L , j ∈ R ( d i s [ i ] + d i s [ j ] ) ⋅ d e p ′ [ l c a ′ ( i , j ) ] + w ⋅ ∑ i ∈ L , j ∈ R d i s ′ ( i , j ) = ∣ R ∣ ⋅ ∑ i ∈ L d i s [ i ] ⋅ d e p ′ [ i ] + ∣ L ∣ ⋅ ∑ j ∈ R d i s [ j ] ⋅ d e p ′ [ j ] + ∑ i ∈ L d i s [ i ] ⋅ ∑ j ∈ R d e p ′ [ j ] + ∑ i ∈ L d e p ′ [ i ] ⋅ ∑ j ∈ R d i s [ j ] − 2 ⋅ ∑ i ∈ L , j ∈ R ( d i s [ i ] + d i s [ j ] ) ⋅ d e p ′ [ l c a ′ ( i , j ) ] + w ⋅ ∑ i ∈ L , j ∈ R d i s ′ ( i , j ) \begin{aligned} ANS &= \sum_{i \in L,j \in R}{(dis(i,u) + dis(j,v) + w)\cdot dis'(i,j)}\\ &= \sum_{i \in L,j\in R}{(dis[i] + dis[j])\cdot dis'(i,j)} + w\cdot \sum_{i \in L,j\in R}{dis'(i,j)}\\ &= \sum_{i \in L,j\in R}{(dis[i] + dis[j])\cdot (dep'[i] + dep'[j] - 2 \cdot dep'[lca'(i,j)])} + w\cdot \sum_{i \in L,j\in R}{dis'(i,j)}\\ &= \sum_{i \in L,j\in R}{(dis[i] + dis[j])\cdot (dep'[i] + dep'[j])} - 2 \cdot\sum_{i \in L,j \in R}{(dis[i] + dis[j])\cdot dep'[lca'(i,j)]} + w\cdot \sum_{i \in L,j\in R}{dis'(i,j)}\\ &= \sum_{i \in L,j\in R}{(dis[i] \cdot dep'[i] + dis[j] \cdot dep'[j])} + \sum_{i \in L,j \in R}{(dis[i] \cdot dep'[j] + dis[j] \cdot dep'[i])}\\ &\ \ \ \ \ \ - 2 \cdot\sum_{i \in L,j \in R}{(dis[i] + dis[j])\cdot dep'[lca'(i,j)]} + w\cdot \sum_{i \in L,j\in R}{dis'(i,j)}\\ &= |R| \cdot\sum_{i \in L}{dis[i] \cdot dep'[i]} + |L| \cdot \sum_{j \in R}{dis[j] \cdot dep'[j]} + \sum_{i \in L}{dis[i]} \cdot \sum_{j \in R}{dep'[j]}+ \sum_{i \in L}{dep'[i]} \cdot \sum_{j \in R}{dis[j]}\\ &\ \ \ \ \ \ - 2 \cdot\sum_{i \in L,j \in R}{(dis[i] + dis[j])\cdot dep'[lca'(i,j)]} + w\cdot \sum_{i \in L,j\in R}{dis'(i,j)} \end{aligned} ANS=i∈L,j∈R∑(dis(i,u)+dis(j,v)+w)⋅dis′(i,j)=i∈L,j∈R∑(dis[i]+dis[j])⋅dis′(i,j)+w⋅i∈L,j∈R∑dis′(i,j)=i∈L,j∈R∑(dis[i]+dis[j])⋅(dep′[i]+dep′[j]−2⋅dep′[lca′(i,j)])+w⋅i∈L,j∈R∑dis′(i,j)=i∈L,j∈R∑(dis[i]+dis[j])⋅(dep′[i]+dep′[j])−2⋅i∈L,j∈R∑(dis[i]+dis[j])⋅dep′[lca′(i,j)]+w⋅i∈L,j∈R∑dis′(i,j)=i∈L,j∈R∑(dis[i]⋅dep′[i]+dis[j]⋅dep′[j])+i∈L,j∈R∑(dis[i]⋅dep′[j]+dis[j]⋅dep′[i]) −2⋅i∈L,j∈R∑(dis[i]+dis[j])⋅dep′[lca′(i,j)]+w⋅i∈L,j∈R∑dis′(i,j)=∣R∣⋅i∈L∑dis[i]⋅dep′[i]+∣L∣⋅j∈R∑dis[j]⋅dep′[j]+i∈L∑dis[i]⋅j∈R∑dep′[j]+i∈L∑dep′[i]⋅j∈R∑dis[j] −2⋅i∈L,j∈R∑(dis[i]+dis[j])⋅dep′[lca′(i,j)]+w⋅i∈L,j∈R∑dis′(i,j)
每个求和式都是可以直接DP的。。。所以用点集 L ∪ R L \cup R L∪R在 T ′ T' T′上建虚树做01DP即可。
复杂度大概。。 O ( n l o g 2 n ) O(nlog^2n) O(nlog2n)。。那么问题来了。。为什么不用点分呢
#pragma GCC optimize(3)
#include