SPOJ Problem Set (classical)SPOJ 913. Query on a tree IIProblem code: 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
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题目大意：给定一颗有边权的树，有两种操作，DIST(i,j)操作询问i和j节点之间的距离，KTH(i,j,k)操作询问i和j节点之间第k个节点的编号。
解题思路：利用树上的倍增就可以搞定。每个节点都保存它的第2^i的父亲。对于DIST询问，只要利用倍增求出lca，然后减一减就好了。对于KTH询问，先求出lca，然后判断是第一个点到lca的路径上还是第二个点到lca的路径上。哎算是水题，不过第一次用倍增，RE好久。
#include <stdio.h>
#include <string.h>
#include <vector>
#define clr(a,b) memset(a,b,sizeof(a))
using namespace std;
const int N=200005;
int n,eid;
int head[N],ed[N<<1],val[N<<1],nxt[N<<1];
vector<int>fa[N];
int sta[N],top,dep[N],dis[N];
void addedge(int s,int e,int v){
ed[eid]=e;val[eid]=v;nxt[eid]=head[s];head[s]=eid++;
}
void dfs(int s,int f,int d,int ds){
fa[s].clear();int k=1;dep[s]=d;dis[s]=ds;
while(top-k>=0){
fa[s].push_back(sta[top-k]);k*=2;
}
sta[top++]=s;
for(int i=head[s];~i;i=nxt[i]){
int e=ed[i],v=val[i];
if(e!=f)dfs(e,s,d+1,ds+v);
}
top--;
}
int lca(int a,int b){
if(a==b)return a;
if(dep[b]>dep[a])swap(a,b);
while(dep[a]>dep[b]){
int len=fa[a].size(),le=0,ri=len,mid;
while(mid=(le+ri)>>1,ri>le){
if(dep[fa[a][mid]]>=dep[b])le=mid+1;
else ri=mid;
}
a=fa[a][ri-1];
}
if(a==b)return a;
while(1){
int len=fa[a].size(),le=0,ri=len,mid;
while(mid=(le+ri)>>1,ri>le){
if(fa[a][mid]!=fa[b][mid])le=mid+1;
else ri=mid;
}
if(ri==0)return fa[a][ri];
a=fa[a][ri-1];b=fa[b][ri-1];
}
return a;
}
int kth(int a,int b,int k){
int r=lca(a,b);
if(dep[a]-dep[r]+1>=k){
int u=dep[a]-k+1;
while(1){
if(u==dep[a])return a;
int len=fa[a].size(),le=0,ri=len,mid;
while(mid=(le+ri)>>1,ri>le){
if(dep[fa[a][mid]]>=u)le=mid+1;
else ri=mid;
}
a=fa[a][ri-1];
}
}
else{
int u=k-dep[a]+dep[r]*2-1;
while(1){
if(u==dep[b])return b;
int len=fa[b].size(),le=0,ri=len,mid;
while(mid=(le+ri)>>1,ri>le){
if(dep[fa[b][mid]]>=u)le=mid+1;
else ri=mid;
}
b=fa[b][ri-1];
}
}
}
int main(){
// freopen("/home/axorb/in","r",stdin);
int T;scanf("%d",&T);
while(T--){
eid=0;clr(head,-1);scanf("%d",&n);
for(int i=1;i<n;i++){
int a,b,c;scanf("%d%d%d",&a,&b,&c);
addedge(a,b,c);addedge(b,a,c);
}
top=0;dfs(1,-1,1,0);
// for(int i=1;i<=n;i++)printf("%d %d %d\n",i,fa[i].size(),dep[i]);
char ss[20];
while(scanf("%s",ss),ss[1]!='O')
if(ss[1]=='I'){
int a,b;scanf("%d%d",&a,&b);
int r=lca(a,b);
printf("%d\n",dis[a]+dis[b]-2*dis[r]);
}
else{
int a,b,c;scanf("%d%d%d",&a,&b,&c);
printf("%d\n",kth(a,b,c));
}
puts("");
}
}