loj10134. 「一本通 4.4 练习 1」Dis

思路：

　　树上倍增模板题

#include<cstdio>
#include<iostream>
#include<vector>
using namespace std;
const int maxn = 10010;
inline void qread(int &x){
    x = 0;
    register int ch = getchar();
    while(ch < '0' || ch > '9')    ch = getchar();
    while(ch >= '0' && ch <= '9'){
        x = 10 * x + ch - 48;
        ch = getchar();
    }
}
int n, m;
vector<int> G[maxn];
vector<int> W[maxn];
int llog[maxn];
int deep[maxn];
int f[maxn][30];
int g[maxn];
inline void init(){
    qread(n);
    qread(m);
    for(int i=1; i<n; ++i){
        int x, y, z;
        qread(x), qread(y), qread(z);
        G[x].push_back(y);
        G[y].push_back(x);
        W[x].push_back(z);
        W[y].push_back(z);
    }
    deep[1] = 1;
}
void dfs(int x, int sum = 0){
    g[x] = sum;
    for(int i=0; i<G[x].size(); ++i)
        if(!deep[G[x][i]]){
            f[G[x][i]][0] = x;
            deep[G[x][i]] = deep[x] + 1;
            dfs(G[x][i], sum + W[x][i]);
        }
}
inline void STtree(){
    for(int i=2; i<=n; ++i)
        llog[i] = llog[i >> 1] + 1;
    for(int j = 1; j <= llog[n]; ++j)
        for(int i=1; i<=n; ++i)
            f[i][j] = f[f[i][j-1]][j-1];        
}
inline int LCA(int x, int y){
    if(x == y)     return x;
    if(deep[x] < deep[y])    swap(x, y);
    for(int j = llog[n]; j>=0; --j)
        if(deep[f[x][j]] >= deep[y])
            x = f[x][j];
    if(x == y)     return x;
    for(int j = llog[n]; j>=0; --j)
        if(f[x][j] != f[y][j]){
            x = f[x][j];
            y = f[y][j];
        }
    return f[x][0];            
}
int main(void){
    init();
    dfs(1);
    STtree();
    while(m--){
        int x, y;
        qread(x), qread(y);
        printf("%d\n", g[x] + g[y] - (g[LCA(x, y)] << 1));
    }
}