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≤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(nlogn)，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;}