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.
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
最简单的LCA，用d[u]存u到根节点距离，然后u到v的距离就为他们的d[i]减去2被的祖先的d[i]
1 #include<cstdio>
2 #include<iostream>
3 #include<cstring>
4 using namespace std;
5 const int maxn=80005;
6 int head[maxn],tov[maxn],w[maxn],nextn[maxn];
7 int fa[maxn][16],deep[maxn];
8 int dis[maxn];
9 bool pd[maxn];
10 int tot=0;
11 int xxx;
12 void go(int x,int y,int z)
13 {
14 tot++;
15 nextn[tot]=head[x];
16 head[x]=tot;
17 tov[tot]=y;
18 w[tot]=z;
19 }
20 void dfs(int x)
21 {
22 int v=head[x];
23 while(v)
24 {
25 if(pd[tov[v]]==false)
26 {
27 pd[tov[v]]=true;
28 deep[tov[v]]=deep[x]+1;
29 dis[tov[v]]=dis[x]+w[v];
30 fa[tov[v]][0]=x;
31 int ii=0,po=x;
32 while(fa[po][ii]!=0)
33 {
34 fa[tov[v]][ii+1]=fa[po][ii];
35 po=fa[po][ii];
36 ii++;
37 }
38 dfs(tov[v]);
39 }
40 v=nextn[v];
41 }
42 }
43 int lca(int x,int y)
44 {
45 if(x==y)
46 return x;
47 if(deep[x]<deep[y]){int t=x;x=y;y=t;}
48 int m=deep[x]-deep[y];
49 int ii=0;
50 while(m!=0)
51 {
52 if(m&1==1)
53 x=fa[x][ii];
54 m=(m>>1);
55 ii++;
56 }
57 ii=0;
58 while(x!=y)
59 {
60 if(fa[x][ii]!=fa[y][ii]||(fa[x][ii]==fa[y][ii]&&(ii==0)))
61 {
62 x=fa[x][ii];y=fa[y][ii];
63 ii++;
64 }
65 else ii--;
66 }
67 return x;
68 }
69
70 int main()
71 {
72 scanf("%d",&xxx);
73 for(int i=1;i<=xxx;i++)
74 {
75 memset(head,0,sizeof(head));
76 memset(nextn,0,sizeof(nextn));
77 memset(tov,0,sizeof(tov));
78 memset(w,0,sizeof(w));
79 memset(fa,0,sizeof(fa));
80 memset(deep,0,sizeof(deep));
81 memset(pd,0,sizeof(pd));
82 memset(dis,0,sizeof(dis));
83
84 int n,m;
85 scanf("%d%d",&n,&m);
86 for(int i=1;i<=n-1;i++)
87 {
88 int x,y,z;
89 scanf("%d%d%d",&x,&y,&z);
90 go(x,y,z);go(y,x,z);
91 }
92 deep[1]=1;
93 pd[1]=true;
94 dfs(1);
95 for(int i=1;i<=m;i++)
96 {
97 int x,y;
98 scanf("%d%d",&x,&y);
99 printf("%d
",dis[x]+dis[y]-2*dis[lca(x,y)]);
100 }
101 }
102 return 0;
103 }