POJ 1986 Distance Queries（查询两点距离，LCA）

题目链接：
POJ 1986 Distance Queries
题意：
给一个连通的$n$个节点的树，有$Q$次查询，每次输出两点间距离。
数据范围：$n\leq 40000$
分析：
先用基于RMQ算法的求LCA的方法求出LCA。记$dis[i]$为根节点到$i$节点的距离，那么$u$和$v$之间的距离就是：

$$Ans=dis[u]+dis[v] - 2 * dis[LCA(u,v)]$$
我建单向边wa了，双向边就AC了，不太懂啊。。。
时间复杂度：预处理: $O(n\log n)$，查询： $O(1)$

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
typedef long long ll;
const int MAX_N = 40010;int n, m, total;
int head[MAX_N], in[MAX_N], id[MAX_N], dis[MAX_N];
int vis[MAX_N * 2], depth[MAX_N * 2], dp[MAX_N * 2][20];struct Edge {int v, w, next;
}edge[MAX_N * 2];void AddEdge (int u, int v, int w)
{edge[total].v = v, edge[total].w = w;edge[total].next = head[u];head[u] = total++;
}void dfs(int u, int p, int d, int& k)
{vis[k] = u, id[u] = k;depth[k++] = d;for (int i = head[u]; i != -1; i = edge[i].next) {int v = edge[i].v, w = edge[i].w;if (v == p) continue;dis[v] = dis[u] + w;dfs(v, u, d + 1, k);vis[k] = u;depth[k++] = d;}
}void RMQ(int root)
{int k = 0;dfs(root, -1, 0, k);int mm = k;int e = (int)log2(mm + 1.0);for (int i = 0; i < mm; ++i) dp[i][0] = i;for (int j = 1; j <= e; ++j) {for (int i = 0; i + (1 << j) - 1 < mm; ++i) {int nxt = i + (1 << (j - 1));if (depth[dp[i][j - 1]] < depth[dp[nxt][j - 1]]) {dp[i][j] = dp[i][j - 1]; } else {dp[i][j] = dp[nxt][j - 1];}}}
}int LCA(int u, int v)
{int left = min(id[u], id[v]), right = max(id[u], id[v]);int k = (int)log2(right - left + 1.0);int pos, nxt = right - (1 << k) + 1;if (depth[dp[left][k]] < depth[dp[nxt][k]]) {pos = dp[left][k];} else {pos = dp[nxt][k];}return dis[u] + dis[v] - 2 * dis[vis[pos]];
}void init()
{total = 0;memset(head, -1, sizeof(head));memset(vis, 0, sizeof(head));memset(in, 0, sizeof(in));memset(dis, 0, sizeof(dis));for (int i = 0; i <= n; ++i) { fa[i] = i; }
}int main()
{while (~scanf("%d%d", &n, &m)) {init();for (int i = 0; i < m; ++i) {int u, v, w;char s[10];scanf("%d%d%d%s", &u, &v, &w, s);AddEdge(u, v, w);AddEdge(v, u, w);in[v]++;}int root;for (int i = 1; i <= n; ++i) {if (in[i] == 0) {root = i;break;}}RMQ(root);scanf("%d", &m);while (m--) {int u, v;scanf("%d%d", &u, &v);printf("%d\n", LCA(u, v));}}return 0;
}