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
Source
ECJTU 2009 Spring Contest
树上倍增求其LCA 即可；
钦定一个根节点，计算每个点到其的距离，
那么树上两点的距离即为：
dis [ x ]+ dis [ y ]- 2 * dis [ LCA(x,y) ]；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 1000005
#define inf 0x3f3f3f3f
#define INF 0x7fffffff
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const int mod = 10000007;
#define Mod 20100403
#define sq(x) (x)*(x)
#define eps 1e-10
const int N = 1505;
inline int rd() {
int x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
int edge[maxn], ver[maxn], head[maxn], nxt[maxn];
int dp[maxn >> 1][20];
int deg[maxn >> 1];
int dis[maxn>>1];
int t, cnt;
queue<int>q;
void addedge(int x, int y, int w) {
ver[++cnt] = y; edge[cnt] = w;
nxt[cnt] = head[x]; head[x] = cnt;
}
void bfs() {
q.push(1); dis[1] = 0; deg[1] = 1;
while (!q.empty()) {
int x = q.front(); q.pop();
for (int i = head[x]; i; i = nxt[i]) {
int y = ver[i];
if (deg[y])continue;
deg[y] = deg[x] + 1;
dis[y] = dis[x] + edge[i];
dp[y][0] = x;
for (int j = 1; j <= t; j++)
dp[y][j] = dp[dp[y][j - 1]][j - 1];
q.push(y);
}
}
}
int Lca(int x, int y) {
if (deg[x] > deg[y])swap(x, y);// d[x]<=d[y]
for (int i = t; i >= 0; i--) {
if (deg[dp[y][i]] >= deg[x])y = dp[y][i];
}
if (x == y)return x;
for (int i = t; i >= 0; i--) {
if (dp[x][i] != dp[y][i])x = dp[x][i], y = dp[y][i];
}
return dp[x][0];
}
int main()
{
//ios::sync_with_stdio(false);
int T; rdint(T);
while (T--) {
int n, m; cnt = 0;
rdint(n); rdint(m);
for (int i = 0; i <= n; i++)deg[i] = dis[i] = 0;
ms(head);
t = (int)(log(n) / log(2)) + 1;
n--;
while ((n)--) {
int u, v, w; rdint(u); rdint(v); rdint(w);
addedge(u, v, w); addedge(v, u, w);
}
bfs();
while (m--) {
int x, y; rdint(x); rdint(y);
cout << dis[x] + dis[y] - 2 * dis[Lca(x, y)] << endl;
}
}
}
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 1000005
#define inf 0x3f3f3f3f
#define INF 0x7fffffff
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const int mod = 10000007;
#define Mod 20100403
#define sq(x) (x)*(x)
#define eps 1e-10
const int N = 1505;
inline int rd() {
int x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
int edge[maxn], ver[maxn], head[maxn], nxt[maxn];
int dp[maxn >> 1][20];
int deg[maxn >> 1];
int dis[maxn>>1];
int t, cnt;
queue<int>q;
void addedge(int x, int y, int w) {
ver[++cnt] = y; edge[cnt] = w;
nxt[cnt] = head[x]; head[x] = cnt;
}
void bfs() {
q.push(1); dis[1] = 0; deg[1] = 1;
while (!q.empty()) {
int x = q.front(); q.pop();
for (int i = head[x]; i; i = nxt[i]) {
int y = ver[i];
if (deg[y])continue;
deg[y] = deg[x] + 1;
dis[y] = dis[x] + edge[i];
dp[y][0] = x;
for (int j = 1; j <= t; j++)
dp[y][j] = dp[dp[y][j - 1]][j - 1];
q.push(y);
}
}
}
int Lca(int x, int y) {
if (deg[x] > deg[y])swap(x, y);// d[x]<=d[y]
for (int i = t; i >= 0; i--) {
if (deg[dp[y][i]] >= deg[x])y = dp[y][i];
}
if (x == y)return x;
for (int i = t; i >= 0; i--) {
if (dp[x][i] != dp[y][i])x = dp[x][i], y = dp[y][i];
}
return dp[x][0];
}
int main()
{
//ios::sync_with_stdio(false);
int T; rdint(T);
while (T--) {
int n, m; cnt = 0;
rdint(n); rdint(m);
for (int i = 0; i <= n; i++)deg[i] = dis[i] = 0;
ms(head);
t = (int)(log(n) / log(2)) + 1;
n--;
while ((n)--) {
int u, v, w; rdint(u); rdint(v); rdint(w);
addedge(u, v, w); addedge(v, u, w);
}
bfs();
while (m--) {
int x, y; rdint(x); rdint(y);
cout << dis[x] + dis[y] - 2 * dis[Lca(x, y)] << endl;
}
}
}
