SPOJ-913

Query on a tree II

Time Limit: 433MS

Memory Limit: 1572864KB

64bit IO Format: %lld & %llu

Submit Status

Description

You are given a tree (an undirected acyclic connected graph) with N nodes, and edges numbered 1, 2, 3...N-1. Each edge has an integer value assigned to it, representing its length.

We will ask you to perfrom some instructions of the following form:

• DIST a b : ask for the distance between node a and node b
  or
• KTH a b k : ask for the k-th node on the path from node a to node b

Example:
N = 6 
1 2 1 // edge connects node 1 and node 2 has cost 1 
2 4 1 
2 5 2 
1 3 1 
3 6 2

Path from node 4 to node 6 is 4 -> 2 -> 1 -> 3 -> 6 
DIST 4 6 : answer is 5 (1 + 1 + 1 + 2 = 5) 
KTH 4 6 4 : answer is 3 (the 4-th node on the path from node 4 to node 6 is 3)

Input

The first line of input contains an integer t, the number of test cases (t <= 25). t test cases follow.

For each test case:

• In the first line there is an integer N (N <= 10000)
• In the next N-1 lines, the i-th line describes the i-th edge: a line with three integers a b c denotes an edge between a, b of cost c (c <= 100000)
• The next lines contain instructions "DIST a b" or "KTH a b k"
• The end of each test case is signified by the string "DONE".

There is one blank line between successive tests.

Output

For each "DIST" or "KTH" operation, write one integer representing its result.

Print one blank line after each test.

Example

Input:
1

6
1 2 1
2 4 1
2 5 2
1 3 1
3 6 2
DIST 4 6
KTH 4 6 4
DONE

Output:
5
3

/**
    题意：给一个树，求u->v的距离
                    求u->v的第k个点
    做法：专题是树链划分 但是想想LCA可以求距离  第k个点 要么是u->v的第k个点  要么是第k`个点
**/
#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdio.h>
#include <cmath>
#include <queue>
#include <set>
using namespace std;
const int maxn = ;
const int DEG = ;
int main();
struct Edge
{
    int to;
    int nxt;
    int val;
} edge[maxn * ];
int head[maxn], tot;
int mmap[maxn];
void addedge(int u, int v, int w)
{
    edge[tot].to = v;
    edge[tot].val = w;
    edge[tot].nxt = head[u];
    head[u] = tot++;
}
void init()
{
    tot = ;
    memset(head, -, sizeof(head));
}
int fa[maxn][DEG];
int deg[maxn];
void bfs(int root)
{
    queue<int>que;
    deg[root] = ;
    mmap[root] = ;
    fa[root][] = root;
    que.push(root);
    while(!que.empty())
    {
        int tmp = que.front();
        que.pop();
        for(int i = ; i < DEG; i++) {
            fa[tmp][i] = fa[fa[tmp][i - ]][i - ];
        }
        for(int i = head[tmp]; i != -; i = edge[i].nxt)
        {
            int v = edge[i].to;
            if(v == fa[tmp][]) {
                continue;
            }
            deg[v] = deg[tmp] + ;
            mmap[v] = mmap[tmp] + edge[i].val;
            fa[v][] = tmp;
            que.push(v);
        }
    }
}
int LCA(int u, int v)
{
    if(deg[u] > deg[v]) {
        swap(u, v);
    }
    int hu = deg[u];
    int hv = deg[v];
    int tu = u;
    int tv = v;
    for(int det = hv - hu, i = ; det; det >>= , i++)
        if(det & ) {
            tv = fa[tv][i];
        }
    if(tu == tv) {
        return tu;
    }
    for(int i = DEG - ; i >= ; i--)
    {
        if(fa[tu][i] == fa[tv][i]) {
            continue;
        }
        tu = fa[tu][i];
        tv = fa[tv][i];
    }
    return fa[tu][];
}
bool flag[maxn];
int query(int u, int v, int k)
{
    int root = LCA(u, v);
    int ans ;
    int i, j;
    //  cout << deg[u] << "  " << deg[root] << endl;
    if(deg[u] - deg[root] +  >= k)
    {
        ans = deg[u] - k + ;
        for(i = ; ( << i) <= deg[u]; i++);
        i--;
        for(j = i; j >= ; j--)
        {
            if(deg[u] - ( << j) >= ans)
            {
                u = fa[u][j];
            }
        }
        return u;
    }
    else
    {
        ans = deg[root] + k - (deg[u] - deg[root] + );
        cout << ans << endl;
        for(i = ; ( << i) <= deg[v]; i++);
        i--;
        for(j = i; j >= ; j--)
        {
            if(deg[v] - ( << j) >= ans)
            {
                v = fa[v][j];
            }
        }
        return v;
    }
}
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n;
        scanf("%d", &n);
        int u, v, w;
        memset(flag, false, sizeof(flag));
        init();
        for(int i = ; i < n - ; i++)
        {
            scanf("%d %d %d", &u, &v, &w);
            addedge(u, v, w);
            addedge(v, u, w);
            flag[v] = true;
        }
        int root;
        for(int i = ; i <= n; i++)
        {
            if(!flag[i])
            {
                root = i;
                break;
            }
        }
        bfs(root);
        char ch[];
        int uu, vv, ww;
        while()
        {
            scanf("%s", ch);
            if(strcmp(ch, "DONE") == ) {
                break;
            }
            else if(strcmp(ch, "DIST") == )
            {
                scanf("%d %d", &uu, &vv);
                // cout << deg[uu] << "  " << deg[vv] << endl;
                // cout << LCA(uu, vv) << ".......\n";
                printf("%d\n", mmap[vv] + mmap[uu] -   * mmap[LCA(uu, vv)]);
            }
            else
            {
                scanf("%d %d %d", &uu, &vv, &ww);
                printf("%d\n", query(uu, vv, ww));
            }
        }
    }
    return ;
}