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
題目大意：給出一棵樹，每條邊上有邊權，求x[i] x[i]x[i]到y[i] y[i]y[i]的路徑邊權之和。
思路概括：
LCA模板題。
任意兩點x和y的距離爲x到根節點的距離+ y到根的距離 - 2 * lca(x, y)到根的距離
例如：如果我們要求點4和點7的路徑之和，我們可以先找到他們LCA點2 ，然後就有
dis[4][7] = dis[1][4] + dis[1][7] - 2 * dis[1][2];
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define lt k<<1
#define rt k<<1|1
#define lowbit(x) x&(-x)
#define lson l,mid,lt
#define rson mid+1,r,rt
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned int uint;
typedef unsigned long long ull;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define mem(a, b) memset(a, b, sizeof(a))
#define int ll
const double pi = acos(-1.0);
const double eps = 1e-6;
const double C = 0.57721566490153286060651209;
const ll mod = 1ll << 32;
const int inf = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const uint INF = 0xffffffff;
const int maxn = 4e4 + 5;
struct node
{
int u, v, w, next;
node (){};
node(int u_, int v_, int w_, int next_)
{
u = u_;
v = v_;
w = w_;
next = next_;
}
}edge[maxn * 2];
struct nodd
{
int u, v, id, next;
nodd(){};
nodd(int u_, int v_, int id_, int next_)
{
u = u_;
v = v_;
id = id_;
next = next_;
}
}qu[maxn * 2];
int n, m, ans;
int head[maxn], head1[maxn];
int cnt, cnt1;
int res[maxn][4];
int de[maxn], pre[maxn], fa[maxn];
bool vis[maxn];
void init()
{
cnt = cnt1= 0;
mem(vis, false);
mem(de, 0);
mem(head, -1);
mem(head1, -1);
mem(qu, 0);
mem(edge, 0);
for(int i=1;i<=n;i++)
{
fa[i] = i;
}
}
void add_edge(int u, int v, int w)
{
edge[cnt] = (node){u, v, w, head[u]};
head[u] = cnt++;
}
void qu_edge(int u, int v, int id)
{
qu[cnt1] = (nodd){u, v, id, head1[u]};
head1[u] = cnt1++;
}
int find_fa(int x)
{
if(fa[x] == x) return x;
else return fa[x] = find_fa(fa[x]);
}
void Tarjan(int x)
{
vis[x] = true;
for(int i=head[x];~i;i=edge[i].next)
{
int v = edge[i].v;
if(!vis[v])
{
de[v] = de[x] + edge[i].w;
Tarjan(v);
int r1 = find_fa(x);
int r2 = find_fa(v);
if(r1 != r2)
{
fa[r2] = r1;
}
}
}
for(int i=head1[x]; ~i; i=qu[i].next)
{
int v = qu[i].v;
if(vis[v])
{
res[qu[i].id][2] = find_fa(v);
}
}
}
signed main()
{
int t;
cin >> t;
while(t--)
{
cin >> n >> m;
init();
for(int i=1;i<n;i++)
{
int u, v, w;
cin >> u >> v >> w;
add_edge(u, v, w);
add_edge(v, u, w);
}
for(int i=1;i<=m;i++)
{
int u, v;
cin >> u >> v;
qu_edge(u, v, i);
qu_edge(v, u, i);
res[i][0] = u;
res[i][1] = v;
}
Tarjan(1);
for(int i=1;i<=m;i++)
{
cout << de[res[i][0]] + de[res[i][1]] - 2 * de[res[i][2]] << endl;
}
}
return 0;
}