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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).
The first line of input contains a single integer q (1?≤?q?≤?1?000).
The next q lines contain the information about the events in chronological order. Each event is described in form 1 v u w if it‘s an event when government makes a new rule about increasing the passing fee of all roads on the shortest path from u to v by w dollars, or in form2 v u if it‘s an event when Barnie goes to cuddle from the intersection v to the intersection u.
1?≤?v,?u?≤?1018,?v?≠?u,?1?≤?w?≤?109 states for every description line.
For each event of second type print the sum of passing fee of all roads Barney passes in this event, in one line. Print the answers in chronological order of corresponding events.
7 1 3 4 30 1 4 1 2 1 3 6 8 2 4 3 1 6 1 40 2 3 7 2 2 4
94 0 32
In the example testcase:
Here are the intersections used:
Source
Codeforces Round #362 (Div. 2)
My Solution
LCA(最近公共祖先) 在有根树中，找出某两个结点u和v最近的公共祖先（或者说，离树根最远的公共祖先）。
类似于这样来访问
inline int LCA(const int &u, const int &v)
{
while (u != v){
if(u < v){ //if(v < u){
swap(u, v);
}
u = u/2; // v/2 is parent of vertex v
}
return u;
}
每次修改的时候直接修改就行
1e18 -> 2^63次所以每次 2*63*q == 1e5
复杂度 O(2*63*n)
此外对于 map<pair<LL, LL>, LL> mpt;//map tree
和 map<LL, map<LL, LL> > mpt;//map tree
前面用 Codeforces上的数据做了测试， 这两种写法时间上是差不多的， 然而在空间复杂度上 前者是后者的一半
所以以后就用map<pair<LL, LL>, LL> mpt;//map tree 来处理 1<= u, v <= 1e18, w 这样节点标号很大的情况好了
如果时间压的比较紧可能可以尝试用
FF 计算机... 柱爷
离散化
vector<pair<int,int>> G[N]
读完后sort每个G[i]
后面二分找
#include <iostream>
#include <cstdio>
#include <map>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 8;
map<pair<LL, LL>, LL> mpt;//map tree
int main()
{
#ifdef LOCAL
freopen("a.txt", "r", stdin);
//freopen("b.txt", "w", stdout);
int T = 1;
while(T--){
#endif // LOCAL
ios::sync_with_stdio(false);
cin.tie(0);
LL q, op, u, v, w, ans;
cin>>q;
while(q--){
cin>>op>>u>>v;
if(op == 1){
cin>>w;
while(u != v){
if(u < v){
swap(u, v);
}
mpt[make_pair(u / 2, u)] += w;
u /= 2;
}
}
else{
ans = 0;
while(u != v){
if(u < v){
swap(u, v);
}
ans += mpt[make_pair(u / 2, u)];
u /= 2;
}
cout<<ans<<endl;
}
}
#ifdef LOCAL
printf("\n");
}
#endif // LOCAL
return 0;
}
Thank you!
------from ProLights
Codeforces Round #362 (Div. 2) C. Lorenzo Von Matterhorn LCA(最近公共祖先)
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原文地址：http://blog.csdn.net/prolightsfxjh/article/details/52295393
