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
n个定点，每个点之间有距离，q次询问，求两点之间的最短路径，和最小生成树有所不同的是n个定点只有n-1个边使之相通，没有连成圆。
用LCA做时，连点之间的距离等于根节点到两点距离之和 减去根节点到点v点的祖先的距离的两倍 即ans[query[u][i].w] = dis[u]+dis[v]-2*dis[findset(v)];
1 #include <iostream>
2 #include <stdio.h>
3 #include <string.h>
4 #include <vector>
5 using namespace std;
6 const int N = 40010;
7 struct Nod{
8 int to, w;
9 Nod(int a = 0, int b = 0){
10 to = a; w = b;
11 }
12 };
13 int ans[N];
14 int par[N],vis[N],mark[N],dis[N];
15 int n,m;
16 vector<Nod> mp[N];
17 vector<Nod> query[N];
18 void init(){
19 for(int i = 0; i <= n; i ++){
20 mp[i].clear();
21 query[i].clear();
22 vis[i] = mark[i] = 0;
23 par[i] = i;
24 }
25 memset(ans,-1,sizeof(ans));
26 }
27 int findset(int x){
28 if(x != par[x]) par[x] = findset(par[x]);
29 return par[x];
30 }
31
32 void dfs(int u){
33 for(int i = 0; i < query[u].size(); i ++){
34 int v = query[u][i].to;
35 if(vis[v]&&ans[query[u][i].w]==-1&&!mark[findset(v)]){
36 ans[query[u][i].w] = dis[u]+dis[v]-2*dis[findset(v)];
37 }
38 }
39 for(int i = 0; i < mp[u].size(); i++){
40 int v = mp[u][i].to;
41 if(!vis[v]){
42 vis[v] = 1;
43 dis[v] = dis[u]+mp[u][i].w;
44 dfs(v);
45 par[v] = u;
46 }
47 }
48 }
49
50 int main(){
51 int t;
52 scanf("%d",&t);
53 while(t--){
54 scanf("%d%d",&n,&m);
55 init();
56 int x, y, z;
57 for(int i = 0; i < n-1; i ++){
58 scanf("%d%d%d",&x,&y,&z);
59 mp[x].push_back(Nod(y,z));
60 mp[y].push_back(Nod(x,z));
61 }
62 for(int i = 1; i <= m; i ++){
63 scanf("%d%d",&x,&y);
64 query[x].push_back(Nod(y,i));
65 query[y].push_back(Nod(x,i));
66 }
67 for(int i = 1; i <= n; i ++){
68 if(!vis[i]){
69 printf("%d+++\n",i);
70 vis[i] = 1;
71 dis[i] = 0;
72 dfs(i);
73 mark[i] = 1;
74 }
75 }
76 for(int i = 1; i <= m; i ++){
77 printf("%d\n",ans[i]);
78 }
79 }
80 return 0;
81 }