SPOJ QTREE2 (LCA - 倍增 在线）

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). ttest 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

思路：
倍增裸题。。。套板子，
求第k个的时候需要处理下，其他没什么。，。
实现代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int M = 2e5+10;
int dist[M],p[M][30],dep[M],head[M];
int cnt1,n;

struct node{
    int to,next,w;
}e[M];

void add(int u,int v,int w){
    e[++cnt1].w=w;e[cnt1].to=v;e[cnt1].next=head[u];head[u]=cnt1;
    e[++cnt1].w=w;e[cnt1].to=u;e[cnt1].next=head[v];head[v]=cnt1;
}

void dfs(int u){
    for(int i = head[u];i != -1;i=e[i].next){
        int v = e[i].to;
        if(v == p[u][0]) continue;
        dep[v] = dep[u] + 1;
        dist[v] = dist[u] + e[i].w;
        p[v][0] = u; //p[i][0]存i的父节点
        dfs(v);
    }
}

void init(){
    for(int j = 1;(1<<j)<=n;j++){
        for(int i = 1;i <= n;i++){
            p[i][j] = p[p[i][j-1]][j-1];
            //cout<<i<<" "<<j<<" "<< p[i][j]<<endl;
        }
    }
}

int lca(int a,int b){
    if(dep[a] > dep[b]) swap(a,b);
    int h = dep[b] - dep[a]; //h为高度差
    for(int i = 0;(1<<i)<=h;i++){  //(1<<i)&f找到h化为2进制后1的位置，移动到相应的位置
        if((1<<i)&h) b = p[b][i];
        //比如h = 5(101),先移动2^0祖先，然后再移动2^2祖先
    }
    //cout<<a<<" "<<b<<endl;
    if(a!=b){
        for(int i = 22;i >= 0;i --){
            if(p[a][i]!=p[b][i]){  //从最大祖先开始，判断a,b祖先，是否相同
                a = p[a][i]; b = p[b][i]; //如不相同，a,b,同时向上移动2^j
            }
        }
        a = p[a][0]; //这时a的father就是LCA
    }
    return a;
}

int kth(int u,int k){
    for(int i = 0;i < 22;i ++)
        if(k >> i&1)
            u = p[u][i];
    return u;
}

int main()
{
    int t,u,v,w,k;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        cnt1 = 0;
        //init();
        memset(head,-1,sizeof(head));
        for(int i = 0;i < n-1;i ++){
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w);
        }
        dfs(1);
        init();
        char s[10];
        while(scanf("%s",s)!=EOF){
            if(s[1]=='O') break;
            scanf("%d%d",&u,&v);
            int num = lca(u,v);
            if(s[1]=='I'){
                printf("%d\n",dist[u]+dist[v]-2*dist[num]);
            }
            if(s[1]=='T'){
                scanf("%d",&k);
                int x = dep[u] - dep[num];
                if(x + 1 >= k)
                    printf("%d\n",kth(u,k-1));
                else printf("%d\n",kth(v,dep[v]+dep[u]-2*dep[num]+1-k));
            }
        }
    }
    return 0;
}