倍增法求 LCA

倍增算法可以在线求树上两个点的LCA，时间复杂度为nlogn

预处理：通过dfs遍历，记录每个节点到根节点的距离dist[u]，深度d[u]
并求出树上每个节点i的2^j祖先f[i][j]

求最近公共祖先，根据两个节点的的深度，如不同，向上调整深度大的节点，使得两个节点在同一层上，如果正好是祖先结束，否则，将连个节点同时上移，查询最近公共祖先。

#include <cstdio>#include <cstdlib>#include <cstring>#include <ctime>#include <cmath>#include <iostream>#include <algorithm>using namespace std;struct node {     node *next;     int where, cost;} *first[100001], a[200001];int n, m, f[100001][21], l, dist[100001], D[100001], c[100001];inline void makelist(int x, int y, int z) {     a[++l].where = y; a[l].cost = z;     a[l].next = first[x]; first[x] = &a[l];}int lca(int x, int y) {     if (D[x] < D[y]) swap(x, y);     int will = D[x] - D[y];     for (int step = 0; will; will >>= 1, ++step)          if (will & 1) x = f[x][step]; //移到同一层     if (x == y) return x;     for (int i = 20; i >= 0; --i) //这里 i 一定要 >=0 ,而不是>=1          if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];     return f[x][0];}int main() {     //freopen("lca.in", "r", stdin);     //freopen("tree.out", "w", stdout);     scanf("%d%d", &n, &m);     memset(first, 0, sizeof(first)); l = 0;     for (int i = 1; i < n; i++) {          int x, y, z;          scanf("%d%d%d", &x, &y, &z);          makelist(x, y, z);          makelist(y, x, z);     }     memset(f, 0, sizeof(f));     memset(dist, 255, sizeof(dist));     dist[1] = 0; c[1] = 1; D[1] = 0;     for (int k = 1, l = 1; l <= k; l++) {//预处理出深度和f[i][j]          int m = c[l];          for (node *x = first[m]; x; x = x->next)               if (dist[x->where] == -1) {                    D[x->where] = D[m] + 1;                    dist[x->where] = dist[m] + x->cost;                    f[x->where][0] = m;                    c[++k] = x->where;               }     }     for (int i = 1; i <= 20; i++)           for (int j = 1; j <= n; j++)               if (f[j][i - 1]) f[j][i] = f[f[j][i - 1]][i - 1];     for (; m--; ) {          int x, y;          scanf("%d%d", &x, &y);          cout<<dist[x]+dist[y]-2*dist[lca(x,y)]<<endl;     }     return 0;}