[HDU]P2586 How far away?[LCA]

Posted 2020-10-13

[HDU]P2586

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18675    Accepted Submission(s): 7274

Problem Description

There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can‘t visit a place twice) between every two houses. Yout task is to answer all these curious people.

Input

First line is a single integer T(T<=10), indicating the number of test cases.
  For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
  Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.

Output

For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.

Sample Input

2

3 2

1 2 10

3 1 15

1 2

2 3

 

2 2

1 2 100

1 2

2 1

Sample Output

10

25

100

100

Source

ECJTU 2009 Spring Contest

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---

这道题就是很裸的LCA，主要是练一下倍增，今天考试一道有关LCA的，我用树剖打竟然T了？（感觉效率有保证，不知道是不是数据问题）

可恶啊，打了很久诶，于是就来学习一下倍增。

代码：

 1 //2017.11.7
 2 //lca
 3 #include<iostream>
 4 #include<cstdio>
 5 #include<cstring>
 6 using namespace std;
 7 inline int read();
 8 namespace lys{
 9     const int N = 4e4 + 7 ;
10     struct edge{
11         int to;
12         int next;
13         int w;
14     }e[N*3];
15     int anc[N][17],dis[N][17],dep[N],pre[N];
16     int n,m,cnt;
17     void swap(int &a,int &b){int t=a;a=b;b=t;}
18     void add(int x,int y,int w){
19         e[++cnt].to=y;e[cnt].next=pre[x];pre[x]=cnt;e[cnt].w=w;
20         e[++cnt].to=x;e[cnt].next=pre[y];pre[y]=cnt;e[cnt].w=w;
21     }
22     void dfs(int node,int deep){
23         dep[node]=deep;
24         int i,v;
25         for(i=1;i<=16;i++) anc[node][i]=anc[anc[node][i-1]][i-1],dis[node][i]=dis[node][i-1]+dis[anc[node][i-1]][i-1];
26         for(i=pre[node];i;i=e[i].next){
27             v=e[i].to;
28             if(v==anc[node][0]) continue ;
29             anc[v][0]=node;
30             dis[v][0]=e[i].w;
31             dfs(v,deep+1);
32         }
33     }
34     int lca(int x,int y){
35         int res=0,i;
36         if(dep[x]<dep[y]) swap(x,y);
37         for(i=16;i>=0;i--)
38             if(dep[y]<=anc[x][i]) res+=dis[x][i],x=anc[x][i];
39         if(x==y) return res;
40         for(i=16;i>=0;i--)
41             if(anc[x][i]!=anc[y][i]) res+=dis[x][i]+dis[y][i],x=anc[x][i],y=anc[y][i];
42         return res+dis[x][0]+dis[y][0];
43     }
44     int main(){
45         memset(pre,0,sizeof pre);
46         int i,u,v,w;
47         n=read(); m=read();
48         cnt=0;
49         for(i=1;i<n;i++){
50             u=read(); v=read(); w=read();
51             add(u,v,w);
52         }
53         dfs(1,1);
54         while(m--){
55             u=read(); v=read();
56             printf("%d\n",lca(u,v));
57         }
58         return 0;
59     }
60 }
61 int main(){
62     int T=read();
63     while(T--) lys::main();
64     return 0;
65 }
66 inline int read(){
67     int kk=0,ff=1;
68     char c=getchar();
69     while(c<‘0‘||c>‘9‘){
70         if(c==‘-‘) ff=-1;
71         c=getchar();
72     }
73     while(c>=‘0‘&&c<=‘9‘) kk=kk*10+c-‘0‘,c=getchar();
74     return kk*ff;
75 }