##### codeforces #362（Div.2）C.Lorenzo Von Matterhorn

Barney lives in NYC. NYC has infinite number of intersections numbered with positive integers starting from 1. There exists a bidirectional road between intersections $i$ and $2i$ and another road between $i$ and $2i + 1$ for every positive integer $i$. You can clearly see that there exists a unique shortest path between any two intersections.

Initially anyone can pass any road for free. But since SlapsGiving is ahead of us, there will $q$ consecutive events happen soon. There are two types of events:
$1.$ Government makes a new rule. A rule can be denoted by integers $v, u$ and $w$. As the result of this action, the passing fee of all roads on the shortest path from $u$ to $v$ increases by $w$ dollars.
$2.$ Barney starts moving from some intersection v and goes to intersection u where there’s a girl he wants to cuddle (using his fake name Lorenzo Von Matterhorn). He always uses the shortest path (visiting minimum number of intersections or roads) between two intersections.
Government needs your calculations. For each time Barney goes to cuddle a girl, you need to tell the government how much money he should pay (sum of passing fee of all roads he passes).

#### Input

The first line of input contains a single integer $q (1 ≤ q ≤ 1 000)$.
$q$ 个事件 $1.u $v$ w$ 从$u$到$v$的费用增加$w$ ; $2.u $v$从$u$到$v$的最短路(经过最少的边的数量)上的花费 给出一棵树，大概最多60层的完全二叉树完全没机会建图了 然后给出类似树链剖分的操作修改两点之间的边权+查询LCA 本题好在给出的一颗树完全二叉树所以只存在唯一的最短路，两点之间的最短路就是他们分别到他们的LCA的距离之和。而这个又是完全二叉树祖父关系很好搞，找LCA的话一直除2只到一样，所以考虑把边权做个维护边和的东西。这里使用映射来存大量的边。 #### Code [cpp] #include <bits/stdc++.h> using namespace std; typedef long long ll; int main() { int q;scanf(“%d”,&q); map<ll,ll> h;//每个点维护它向上的一条边, 非常优雅的操作 while(q–) { ll ans = 0; int op;scanf(“%d”,&op); if(op==1) { ll u,v,w;scanf(“%I64d%I64d%I64d”,&u,&v,&w); while(u!=v) //维护边和 if(u>v) h[u]+=w, u/=2; else h[v]+=w, v/=2; } else { ll u,v;scanf(“%I64d%I64d”,&u,&v); while(u!=v)//累加边和 if(u>v) ans += h[u], u/=2; else ans += h[v], v/=2; printf(“%I64d$n”,ans);
}
}
}
[/cpp]

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