[Codevs2370]小机房的树 树上倍增

题意

给出一颗带权树，多次询问两点最小距离

思路

倍增复习题

树上最短距离为两点到其 \(LCA\) 距离，倍增时顺便求和即可

#include <cstdio>

const int maxp = 5e5 + 10, maxe = (7e5 + 1) * 2;
int n, m, edge[maxe], Next[maxe], head[maxp], ver[maxe];
int cnt, fa[maxp][31], dep[maxp], cost[maxp][31];

inline void addline(int x, int y, int z) {
  ver[++cnt] = y, edge[cnt] = z, Next[cnt] = head[x], head[x] = cnt;
}

inline void swap(int &x, int &y) {
  int t = x; x = y; y = t;
}

inline void dfs(int o, int fno) {
  dep[o] = dep[fa[o][0] = fno] + 1;
  for (int i = 1; i < 31; ++ i) {
    fa[o][i] = fa[fa[o][i - 1]][i - 1];
    cost[o][i] = cost[fa[o][i - 1]][i - 1] + cost[o][i - 1];
  }
  for (int i = head[o], v; i; i = Next[i]) {
    if ((v = ver[i]) == fno) continue;
    cost[v][0] = edge[i];
    dfs(v, o);
  }
}

int lca(int x, int y) {
  if (dep[x] > dep[y]) swap(x, y);
  int tmp = dep[y] - dep[x], ans = 0;
  for (int j = 0; tmp; ++ j, tmp >>= 1)
    if (tmp & 1) ans += cost[y][j], y = fa[y][j];
  if (x == y) return ans;
  for (int j = 30; ~j && y != x; -- j) {
    if (fa[x][j] != fa[y][j]) {
      ans += cost[x][j] + cost[y][j];
      x = fa[x][j];
      y = fa[y][j];
    }
  }
  ans += cost[x][0] + cost[y][0];
  return ans;
}

int main() {
  scanf("%d", &n);
  for (int i = 1, x, y, z; i < n; ++ i) {
    scanf("%d %d %d", &x, &y, &z); ++x, ++y;
    addline(x, y, z), addline(y, x, z);
  }
  dfs(1, 0);
  scanf("%d", &m);
  for (int i = 1, x, y; i <= m; ++ i) {
    scanf("%d %d", &x, &y);
    printf("%d\n", lca(x + 1, y + 1));
  }
}