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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:
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)
The first line of input contains an integer t, the number of test cases (t <= 25). t test cases follow.
For each test case:
There is one blank line between successive tests.
For each "DIST" or "KTH" operation, write one integer representing its result.
Print one blank line after each test.
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
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\n",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\n",get(u,d));
}
}
}
}
【SPOJ QTREE2】QTREE2 - Query on a tree II（LCA）
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原文地址：http://www.cnblogs.com/flipped/p/5827406.html
