M - Lorenzo Von Matterhorn (二叉树 )

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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 uwhere 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

Output

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.

Example

Input

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

Output

94
0
32

Note

In the example testcase:

Here are the intersections used:

1. Intersections on the path are 3, 1, 2 and 4.
2. Intersections on the path are 4, 2 and 1.
3. Intersections on the path are only 3 and 6.
4. Intersections on the path are 4, 2, 1 and 3. Passing fee of roads on the path are 32, 32 and 30 in order. So answer equals to 32?+?32?+?30?=?94.
5. Intersections on the path are 6, 3 and 1.
6. Intersections on the path are 3 and 7. Passing fee of the road between them is 0.
7. Intersections on the path are 2 and 4. Passing fee of the road between them is 32 (increased by 30 in the first event and by 2 in the second).

解法:这是网上看到的

因为要计算u->v的权值之和，我们可以把权值放在v中，由于题目中给定的u、v特性，我们可以从最后一个v开始倒回来每次除以2，然后把权值加起来就好了，注意输入的区间大小值

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <map>
 4 using namespace std;
 5 #define LL long long
 6 map<LL, LL>node;
 7 int q, com;
 8 LL u, v, w;
 9 int main()
10 {
11     while(~scanf("%d", &q)){
12         while(q--){
13             scanf("%d", &com);
14             if(com==1){
15                 scanf("%lld%lld%lld", &u, &v, &w);
16                 while(u!=v){
17                     if(u>v){
18                         node[u]+=w;
19                         u/=2;
20                     }
21                     else{
22                         node[v]+=w;
23                         v/=2;
24                     }
25                 }
26             }
27             else{
28                 scanf("%lld%lld", &u, &v);
29                 LL ans=0;
30                 while(u!=v){
31                     if(u>v){
32                         ans+=node[u];
33                         u/=2;
34                     }
35                     else{
36                         ans+=node[v];
37                         v/=2;
38                     }
39                 }
40                 printf("%lld\n", ans);
41             }
42         }
43     }
44     return 0;
45 }

 

M - Lorenzo Von Matterhorn (二叉树 )

标签：mic   des   star   note   ios   make   注意   intersect   nbsp   

原文地址：http://www.cnblogs.com/a2985812043/p/7224574.html