Title:

https://vjudge.net/problem/SPOJ-QTREE2

Title:

Given a tree with edge weight, there are the following operations:

• DIST a b: Find the distance from a to B
• KTH a B k: Find the k-th node on the path from a to b, A is the first

Use DONE as the closing flag of the query

Train of thought:

The first operation, any lca algorithm can be found. Considering the second operation, the simplest method is to look up step by step until the k. It can be found that looking up with doubling idea is log level, so doubling lca is used.

#include <bits/stdc++.h>

using namespace std;

const int N = 10000 + 10, INF = 0x3f3f3f3f;

struct edge
{
    int to, cost, next;
}g[N*2];

int cnt, head[N];
int dep[N], dis[N], fat[N][20];

void init()
{
    cnt = 0;
    memset(head, -1, sizeof head);
}
void add_edge(int v, int u, int cost)
{
    g[cnt].to = u, g[cnt].cost = cost, g[cnt].next = head[v], head[v] = cnt++;
}
void dfs(int v, int fa, int d, int val)
{
    dep[v] = d, fat[v][0] = fa, dis[v] = val;
    for(int i = head[v]; ~i; i = g[i].next)
    {
        int u = g[i].to;
        if(u == fa) continue;
        dfs(u, v, d+1, val + g[i].cost);
    }
}
void lca_init(int n)
{
    for(int j = 1; (1<<j) <= n; j++)
        for(int i = 1; i <= n; i++)
            fat[i][j] = fat[fat[i][j-1]][j-1];
}
int lca(int v, int u)
{
    if(dep[v] < dep[u]) swap(v, u);
    int d = dep[v] - dep[u];
    for(int i = 0; (d>>i) != 0; i++)
        if((d>>i) & 1) v = fat[v][i];
    if(v == u) return v;
    for(int i = 18; i >= 0; i--)
        if(fat[v][i] != fat[u][i]) v = fat[v][i], u = fat[u][i];
    return fat[v][0];
}
int query_kth(int v, int u, int k)
{
    int t = lca(v, u);
    int len = dep[v] - dep[t] + 1;
    if(k <= len)
    {
        --k;
        for(int i = 0; (k>>i) != 0; i++)
            if((k>>i) & 1) v = fat[v][i];
        return v;
    }
    else
    {
        k = dep[u] - dep[t] - (k - len);
        for(int i = 0; (k>>i) != 0; i++)
            if((k>>i) & 1) u = fat[u][i];
        return u;
    }
}
int main()
{
    int t, n;
    scanf("%d", &t);
    while(t--)
    {
        init();
        scanf("%d", &n);
        int a, b, c;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d%d%d", &a, &b, &c);
            add_edge(a, b, c); add_edge(b, a, c);
        }
        dfs(1, 0, 1, 0);
        lca_init(n);
        char opt[20];
        while(scanf("%s", opt), opt[1] != 'O')
        {
            scanf("%d%d", &a, &b);
            if(opt[0] == 'D')
            {
                int ans = lca(a, b);
                printf("%d\n", dis[a] - 2 * dis[ans] + dis[b]);
            }
            else
            {
                scanf("%d", &c);
                printf("%d\n", query_kth(a, b, c));
            }
        }
    }
    return 0;
}