SPOJ 3978 Distance Query（tarjan求LCA）

The traffic network in a country consists of N cities (labeled with integers from 1 to N) and N-1 roads connecting

the cities. There is a unique path between each pair of different cities, and we know the exact length of each road. 

Write a program that will, for each of the K given pairs of cities, find the length of the shortest and the length

of the longest road on the path between the two cities.

Input

The first line of input contains an integer N, 2 ≤ N ≤ 100 000. Each of the following N-1 lines contains three

integers A, B and C meaning that there is a road of length C between city A and city B. 
The length of each road will be a positive integer less than or equal to 1 000 000. 
The next line contains an integer K, 1 ≤ K ≤ 100 000. Each of the following K lines contains two different

integers D and E – the labels of the two cities constituting one query.

Output

Each of the K lines of output should contain two integers – the lengths from the task description for the

corresponding pair of the cities.

题目大意：给一棵n个点的树，每条边有一个权值，k个询问，问u到v的简单路径中，权值最小和最大分别为多少。

思路：首先要会普通的tarjan求LCA的算法，在合并集合的时候算出每个点到其根节点的最小和最大权值，在求出某一对询问(u, v)的LCA之后，回溯到他们的LCA的时候把LCA的子集都合并到了LCA上，那么u和v分别到LCA的最小最大权值就知道了，再取其中的最小最大值即可。

PS：时间复杂度为O(n+k)

代码（6470MS）：

#include <cstdio>
#include <cstring>
#include <queue>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
#define X first
#define Y second
typedef pair<int, int> PII;
typedef vector<PII> VPII;
typedef vector<int> VI;

const int MAXN = 100010;
const int MAXE = MAXN << 1;
const int INF = 0x7fff7fff;

int head[MAXN], to[MAXE], next[MAXE], cost[MAXE], ecnt;
int n, m, fa[MAXN];

PII edge[MAXN], a[MAXN], ans[MAXN];
VPII query[MAXN];
VI b[MAXN];

bool vis[MAXN];

void init() {
    for(int i = 1; i <= n; ++i) fa[i] = i;
    ecnt = 2;
}

void add_edge(int u, int v, int w) {
    to[ecnt] = v; cost[ecnt] = w; next[ecnt] = head[u]; head[u] = ecnt++;
    to[ecnt] = u; cost[ecnt] = w; next[ecnt] = head[v]; head[v] = ecnt++;
}

int get_set(int x) {
    if(fa[x] == x) return x;
    int ret = get_set(fa[x]);
    edge[x].X = max(edge[x].X, edge[fa[x]].X);
    edge[x].Y = min(edge[x].Y, edge[fa[x]].Y);
    return fa[x] = ret;
}

void LCA(int u, int f) {
    edge[u].X = 0; edge[u].Y = INF;
    for(int p = head[u]; p; p = next[p]) {
        int &v = to[p];
        if(v == f) continue;
        LCA(v, u);
        edge[v].X = edge[v].Y = cost[p];
        fa[v] = u;
    }
    vis[u] = true;
    for(VPII::iterator it = query[u].begin(); it != query[u].end(); ++it)
        if(vis[it->X]) b[get_set(it->X)].push_back(it->Y);
    for(VI::iterator it = b[u].begin(); it != b[u].end(); ++it) {
        int id = *it, u = a[id].X, v = a[id].Y;
        get_set(u); get_set(v);
        ans[id] = make_pair(max(edge[u].X, edge[v].X), min(edge[u].Y, edge[v].Y));
    }
}

int main() {
    scanf("%d", &n);
    init();
    for(int i = 1; i < n; ++i) {
        int u, v, w;
        scanf("%d%d%d", &u, &v, &w);
        add_edge(u, v, w);
    }
    scanf("%d", &m);
    for(int i = 1; i <= m; ++i) {
        scanf("%d%d", &a[i].X, &a[i].Y);
        query[a[i].X].push_back(make_pair(a[i].Y, i));
        query[a[i].Y].push_back(make_pair(a[i].X, i));
    }
    LCA(1, 0);
    for(int i = 1; i <= m; ++i) printf("%d %d
", ans[i].Y, ans[i].X);
}