题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2586
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.
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
Source
题意：
一个村庄有 n 个房子和 n-1 条双向路，每两个房子之间都有一条简单路径。
如今有m次询问。求两房子之间的距离。
PS:
能够用LCA来解，首先找到u, v 两点的lca，然后计算一下距离值就能够了。
计算方法是。记下根结点到随意一点的距离dis[i]，
这样ans = dis[u] + dis[v] - 2 * dis[lca(v, v)]了。
这题要用c++交。G++会爆栈！
代码例如以下：看别人的模板（tarjan 离线）
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 40047
#define maxm 247 struct node
{
int to,w,next;
} edge[maxn*2]; int n, m; //点数，询问次数
int head[maxn];
int k;
int fa[maxn]; //父亲结点
int dis[maxn]; //到根节点距离
int vis[maxn]; //是否訪问过
int s[maxm]; //询问起点
int e[maxm]; //询问终点
int lca[maxm]; //LCA(s,e) 近期公共祖先 int find(int x)
{
if(fa[x]!=x) return fa[x]=find(fa[x]);
return fa[x];
} void init()
{
k = 1;
memset(head,0,sizeof(head));
memset(dis,0,sizeof(dis));
memset(vis,0,sizeof(vis));
} void add(int u,int v,int w)
{
edge[k].to = v;
edge[k].w = w;
edge[k].next = head[u];
head[u] = k++;
} void tarjan(int u)
{
int i,v;
fa[u] = u;
vis[u] = 1;
for(i = 0; i < m; i++)
{
if(e[i]==u && vis[s[i]])
lca[i] = find(s[i]); //若询问的两点中有一点已被訪问过。则两点的LCA则为这一点的当前父节点
if(s[i]==u && vis[e[i]])
lca[i] = find(e[i]);
}
for(i = head[u]; i; i = edge[i].next)
{
v = edge[i].to;
if(!vis[v]) //若没被訪问过
{
dis[v] = dis[u]+edge[i].w;//更新距离
tarjan(v);
fa[v] = u;//回溯更新父节点
}
}
} int main()
{
int t;
scanf("%d",&t);
while(t--)
{
init();
scanf("%d%d",&n,&m);
int u, v, w;
for(int i = 0; i < n-1; i++)
{
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);
add(v,u,w);
}
for(int i = 0; i < m; i++)
{
scanf("%d%d",&s[i],&e[i]);
}
tarjan(1); for(int i = 0; i < m; i++)
{
printf("%d\n",dis[s[i]]+dis[e[i]]-2*dis[lca[i]]);//两点距离为根节点到两点距离之和-根节点到LCA距离*2
}
}
return 0;
}
（ST在线算法 转）
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
//#pragma comment(linker, "/STACK:102400000,102400000") //不须要申请系统栈
const int N = 40010;
const int M = 25;
int dp[2*N][M]; //这个数组记得开到2*N，由于遍历后序列长度为2*n-1
bool vis[N];
struct edge
{
int u,v,w,next;
} e[2*N];
int tot,head[N];
inline void add(int u ,int v ,int w ,int &k)
{
e[k].u = u;
e[k].v = v;
e[k].w = w;
e[k].next = head[u];
head[u] = k++;
u = u^v;
v = u^v;
u = u^v;
e[k].u = u;
e[k].v = v;
e[k].w = w;
e[k].next = head[u];
head[u] = k++;
}
int ver[2*N],R[2*N],first[N],dir[N];
//ver:节点编号 R：深度 first：点编号位置 dir：距离
void dfs(int u ,int dep)
{
vis[u] = true;
ver[++tot] = u;
first[u] = tot;
R[tot] = dep;
for(int k=head[u]; k!=-1; k=e[k].next)
if( !vis[e[k].v] )
{
int v = e[k].v , w = e[k].w;
dir[v] = dir[u] + w;
dfs(v,dep+1);
ver[++tot] = u;
R[tot] = dep;
}
}
void ST(int n)
{
for(int i=1; i<=n; i++)
dp[i][0] = i;
for(int j=1; (1<<j)<=n; j++)
{
for(int i=1; i+(1<<j)-1<=n; i++)
{
int a = dp[i][j-1] , b = dp[i+(1<<(j-1))][j-1];
dp[i][j] = R[a]<R[b]?a:b;
}
}
}
//中间部分是交叉的。
int RMQ(int l,int r)
{
int k=0;
while((1<<(k+1))<=r-l+1)
k++;
int a = dp[l][k], b = dp[r-(1<<k)+1][k]; //保存的是编号
return R[a]<R[b]?a:b;
} int LCA(int u ,int v)
{
int x = first[u] , y = first[v];
if(x > y) swap(x,y);
int res = RMQ(x,y);
return ver[res];
} int main()
{
//freopen("Input.txt","r",stdin);
//freopen("Out.txt","w",stdout);
int cas;
scanf("%d",&cas);
while(cas--)
{
int n,q,num = 0;
scanf("%d%d",&n,&q);
memset(head,-1,sizeof(head));
memset(vis,false,sizeof(vis));
for(int i=1; i<n; i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
add(u,v,w,num);
}
tot = 0;
dir[1] = 0;
dfs(1,1);
/*printf("节点ver "); for(int i=1; i<=2*n-1; i++) printf("%d ",ver[i]); cout << endl;
printf("深度R "); for(int i=1; i<=2*n-1; i++) printf("%d ",R[i]); cout << endl;
printf("首位first "); for(int i=1; i<=n; i++) printf("%d ",first[i]); cout << endl;
printf("距离dir "); for(int i=1; i<=n; i++) printf("%d ",dir[i]); cout << endl;*/
ST(2*n-1);
while(q--)
{
int u,v;
scanf("%d%d",&u,&v);
int lca = LCA(u,v);
printf("%d\n",dir[u] + dir[v] - 2*dir[lca]);
}
}
return 0;
}