spoj 913 Query on a tree II (倍增lca)

Query on a tree II

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
orKTH a b k : ask for the k-th node on the path from node a to node bExample:
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 2Path 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

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <time.h>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
#define lson(x) ((x<<1))
#define rson(x) ((x<<1)+1)
using namespace std;
typedef long long ll;
const int N=1e4+50;
const int M=N*N+10;
int n,m,k,tot=0;
int fa[2*N][20],head[N*2],dis[N*2],dep[N*2];
struct man{
int to,next,w;
}edg[2*N];
void add(int u,int v,int w){
edg[tot].to=v;edg[tot].next=head[u];edg[tot].w=w;head[u]=tot++;
}
void init(){
met(head,-1);met(fa,0);met(dis,0);met(dep,0);
tot=0;
}
void dfs(int u,int f){
fa[u][0]=f;
for(int i=1;i<20;i++){
fa[u][i]=fa[fa[u][i-1]][i-1];
}
for(int i=head[u];i!=-1;i=edg[i].next){
int v=edg[i].to;
if(v!=f){
dis[v]=dis[u]+edg[i].w;
dep[v]=dep[u]+1;
dfs(v,u);
}
}
}
int LCA(int u,int v){
int U=u,V=v;
if(dep[u]<dep[v])swap(u,v);
for(int i=19;i>=0;i--){
if(dep[fa[u][i]]>=dep[v]){
u=fa[u][i];
}
}
if(u==v)return (u);
for(int i=19;i>=0;i--){
if(fa[u][i]!=fa[v][i]){
u=fa[u][i];v=fa[v][i];
}
}
return (fa[u][0]);
}
int find_kth(int u,int v,int lca,int k){
k--;
if(dep[u]-dep[lca]<k){
k=dep[u]-dep[lca]*2+dep[v]-k;
u=v;
}
for(int i=0;i<20;i++){
if(k&(1<<i)){          //注意这里
u=fa[u][i];
}
}
return u;
}
void solve(){
int u,v,val;
scanf("%d",&n);
for(int i=1;i<n;i++){
scanf("%d%d%d",&u,&v,&val);
add(u,v,val);add(v,u,val);
}
dep[1]=1;
dfs(1,0);
char str[20];
while(1){
scanf("%s",str);
if(str[0]=='D'&&str[1]=='I'){
scanf("%d%d",&u,&v);
int lca=LCA(u,v);
printf("%d\n",dis[u]+dis[v]-2*dis[lca]);
}
else if(str[0]=='K'){
scanf("%d%d%d",&u,&v,&k);
int lca=LCA(u,v);
printf("%d\n",find_kth(u,v,lca,k));
}
else break;
}
}
int main(){
int t;
scanf("%d",&t);
while(t--){
init();
solve();
}
return 0;
}