【SPOJ QTREE2】QTREE2

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

 Submit solution!

 先以一点为根做dfs，计算每个点的深度、父亲、离根的距离，倍增法找出LCA，两点的距离就能算出来了。

路径上的第K个值则先判断LCA到起点的深度差是否大于k，是则在起点到LCA的路径上，否则在LCA到终点的路径上。

#include <iostream>
#include <cstdio>
#include <cstring>
#define N 10005
using namespace std;
struct edge{
    int to,next,w;
}e[N<<1];
int head[N],cnt;
void add(int u,int v,int w){
    e[++cnt]=(edge){v,head[u],w};
    head[u]=cnt;
}
int n;
int deep[N],dis[N],p[N][30];
void dfs(int x,int fa){
    deep[x]=deep[fa]+1;
    p[x][0]=fa;
    for(int i=0;p[x][i];i++)
        p[x][i+1]=p[p[x][i]][i];//由x的2^i祖先和祖先的2^i祖先算出x的2^(i+1)祖先
    for(int i=head[x];i;i=e[i].next){
        int to=e[i].to;
        if(to==fa)continue;
        dis[to]=dis[x]+e[i].w;
        dfs(to,x);
    }
}
int lca(int a,int b){
    if(deep[a]<deep[b])swap(a,b);
    for(int j=14;j>=0;j--)
    if(deep[a]-(1<<j)>=deep[b])//将a上移到和b深度一样
        a=p[a][j];
    if(a==b)return a;
    for(int j=14;j>=0;j--)
        if(p[a][j]&&p[a][j]!=p[b][j]){//a、b一起上移
            a=p[a][j];
            b=p[b][j];
        }
    return p[a][0];
}
int get(int u,int d){//u上升到d深度
    for(int j=14;j>=0;j--)
    if(deep[u]-(1<<j)>=d)
        u=p[u][j];
    return u;
}
int main() {
    int t;
    cin>>t;
    while(t--){
        memset(head,0,sizeof head);
        memset(p,0,sizeof p);//这个要清空！！
        cnt=0;
        cin>>n;
        for(int i=1;i<n;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w);
            add(v,u,w);
        }
        //deep[0]=0;dis[1]=0;这可以不写
        dfs(1,0);
        char op[100];
        while(1){
            scanf("%s",op);
            if(op[1]=='O')break;
            if(op[1]=='I'){
                int u,v;
                scanf("%d%d",&u,&v);
                printf("%d
",dis[u]+dis[v]-2*dis[lca(u,v)]);
            }else{ 
                int u,v,k;
                scanf("%d%d%d",&u,&v,&k);
                int fa=lca(u,v),d;
                if(deep[u]-deep[fa]<k){
                    d=deep[fa]+k-(deep[u]-deep[fa]+1);
                    u=v;
                }
                else d=deep[u]-k+1;//需要的深度
                printf("%d
",get(u,d));
            }
        }
    }
}