HDU - 2586 How far away ？

Posted 2020-10-04 西北会法语

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.InputFirst 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.OutputFor 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
LCA倍增

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 
 5 using namespace std;
 6 
 7 const int maxn=40005;
 8 const int LOG = 20;
 9 int par[maxn][LOG],dep[maxn];
10 int head[maxn],dis[maxn];
11 int cnt=0;
12 
13 struct EDGE
14 {
15     int v,w,next;
16 }edge[maxn*2];
17 
18 void addedge(int u,int v,int w)
19 {
20     edge[cnt].v=v;
21     edge[cnt].w=w;
22     edge[cnt].next=head[u];
23     head[u]=cnt++;
24 }
25 
26 void dfs(int u,int fa,int depth){
27     dep[u]=depth;
28     if(u==1)
29     { 
30        for(int i=0;i<LOG;i++) par[u][i]=u;
31     }
32         else
33         {
34             par[u][0]=fa; 
35             for(int i=1;i<LOG;i++) par[u][i]=par[par[u][i-1]][i-1];
36         }
37     for(int i=head[u];i!=-1;i=edge[i].next)
38     {
39         int q=edge[i].v;
40         if(q==fa) continue;
41         dis[q]=dis[u]+edge[i].w;
42         dfs(q,u,depth+1);
43     }
44 }
45 
46 int up(int x,int step){ 
47     for(int i=0;i<LOG;i++) 
48         if(step&(1<<i))
49             x=par[x][i];
50     return x;
51 }
52 
53 int lca(int u,int v){
54     if(dep[u]<dep[v]) 
55         swap(u,v);   
56     u=up(u,dep[u]-dep[v]);      
57     if(u==v) return u;    
58     for(int i=LOG-1;i>=0;i--)
59     {  
60         if(par[u][i]!=par[v][i])
61         {
62             u=par[u][i];
63             v=par[v][i];
64         }
65     }
66     return par[u][0];
67 }
68 
69 int main()
70 {
71     int T,n,m;
72     cin>>T;
73     while(T--)
74     {
75         cin>>n>>m;
76         cnt=0;
77         memset(head,-1,sizeof(head));
78         int x,y,z;
79         for(int i=0;i<n-1;i++)
80         {
81             scanf("%d%d%d",&x,&y,&z);
82             addedge(x,y,z);
83             addedge(y,x,z);
84         }
85         dis[1]=0;
86         dfs(1,-1,0);
87         for(int i=0;i<m;i++)
88         {
89             scanf("%d%d",&x,&y);
90             int ans=dis[x]+dis[y]-2*dis[lca(x,y)];
91             printf("%d\n",ans);
92         }
93     }
94     
95     return 0;
96 }