POJ 1986 Distance Queries——LCA

题意：

树上两点间的最近距离

思路：

Tarjan求LCA， dis[u]+dis[v]-dis[lca(u, v)]即可

边表居然比vector快5倍。。。绝了

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 1e5 + 10;
char ch[100];
int n, m, q;
int tot1, tot2, head1[maxn], head2[maxn];
struct Edge { int to, x, next; } edge1[maxn], edge2[maxn];
void addedge1(int u, int v, int dis) {
    edge1[++tot1].to = v; edge1[tot1].x = dis; edge1[tot1].next = head1[u]; head1[u] = tot1;
}
void addedge2(int u, int v, int id) {
    edge2[++tot2].to = v; edge2[tot2].x = id; edge2[tot2].next = head2[u]; head2[u] = tot2;
}
struct Ans { int u, v, lca; } ans[maxn];
int dis[maxn], vis[maxn];
int fa[maxn];
void init() {
    tot1 = tot2 = 0;
    for (int i = 1; i <= n; i++) head1[i] = head2[i] = -1;
    dis[1] = 0;
    for (int i = 1; i <= n; i++) vis[i] = 0, fa[i] = i;
}
int query(int x) { return (fa[x] == x ? x : fa[x] = query(fa[x])); }
void dfs(int u, int p) {
    vis[u] = 1;
    for (int i = head1[u]; ~i; i = edge1[i].next) {
        int v = edge1[i].to, d = edge1[i].x;
        if (v == p) continue;
        dis[v] = dis[u] + d;
        dfs(v, u);
        fa[v] = u;
    }
    for (int i = head2[u]; ~i; i = edge2[i].next) {
        int v = edge2[i].to, id = edge2[i].x;
        if (!vis[v]) continue;
        ans[id].lca = query(v);
    }
}
int main() {
    while (~scanf("%d%d", &n, &m)) {
        init();
        int u, v, d;
        while (m--) {
            scanf("%d%d%d%s", &u, &v, &d, ch);
            addedge1(u, v, d); addedge1(v, u, d);
        }
        scanf("%d", &q);
        for (int i = 1; i <= q; i++) {
            scanf("%d%d", &u, &v);
            addedge2(u, v, i); addedge2(v, u, i);
            ans[i].u = u, ans[i].v = v, ans[i].lca = 1;
        }
        dfs(1, -1);
        for (int i = 1; i <= q; i++) {
            printf("%d\n", dis[ans[i].u] + dis[ans[i].v] - 2*dis[ans[i].lca]);
        }
    }
    return 0;
}