LCA

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.

InputFirst 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.OutputFor 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

题目分析 ： 给出一些点之间的长度，再给出一些询问，查任意两点间的距离。
思路分析 ： 倍增LCA裸题
代码示例 ：

#define ll long long
const int maxn = 4e4+5;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;

int n, m, N;
struct node
{
    int to, cost;
    node(int _to = 0, int _cost = 0):to(_to),cost(_cost){}
};
vector<node>ve[maxn];
int dep[maxn];
int grand[maxn][30], gw[maxn][30];

void dfs(int x, int fa){
    for(int i = 1; i <= N; i++){
        grand[x][i] = grand[grand[x][i-1]][i-1];
        gw[x][i] = gw[x][i-1] + gw[grand[x][i-1]][i-1];
    }
    for(int i = 0; i < ve[x].size(); i++){
        int to = ve[x][i].to;
        int cost = ve[x][i].cost;
        
        if (to == fa) continue;
        dep[to] = dep[x] + 1;
        grand[to][0] = x;
        gw[to][0] = cost;
        dfs(to, x);
    }
}

int lca(int a, int b){
    // a 是在 b 的上面的
    if (dep[a] > dep[b]) swap(a, b);
    int ans = 0;
    
    for(int i = N; i >= 0; i--){
        if (dep[a] < dep[b] && dep[grand[b][i]] >= dep[a]){
            ans += gw[b][i];
            b = grand[b][i];
        }
    }
    // a, b 在同一层后
    for(int i = N; i >= 0; i--){
        if (grand[a][i] != grand[b][i]) { 
            ans += gw[a][i];
            ans += gw[b][i];
            a = grand[a][i], b = grand[b][i];
        }
    } 
    if (a != b) {
        ans += gw[a][0];
        ans += gw[b][0];
    }
    return ans;
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int t;
    int a, b, c;

    cin >> t; 
    while(t--){
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i++) ve[i].clear();
        for(int i = 1; i < n; i++){
            scanf("%d%d%d", &a, &b, &c);
            ve[a].push_back(node(b, c));
            ve[b].push_back(node(a, c));
        }    
        memset(grand, 0, sizeof(grand));
        memset(gw, 0, sizeof(gw));
        dep[1] = 0; 
        N = floor(log(n)/log(2)); // 求出最大的2^k = N中的 k
        dfs(1, 1);
        for(int i = 1; i <= m; i++){
            scanf("%d%d", &a, &b);
            printf("%d
", lca(a, b));
        }
    }
    return 0;
}

/*
5
8 100
1 2 1
2 4 1
2 5 1
2 6 1
1 3 1
3 7 1
1 8 1
*/