| Time Limit: 433MS | Memory Limit: 1572864KB | 64bit IO Format: %lld & %llu |
Description
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
Hint
| Added by: | Thanh-Vy Hua |
| Date: | 2006-08-27 |
| Time limit: | 0.433s |
| Source limit: | 15000B |
| Memory limit: | 1536MB |
| Cluster: | Cube (Intel G860) |
| Languages: | All except: ERL JS NODEJS PERL 6 VB.net |
| Resource: | Special thanks to Ivan Krasilnikov for his alternative solution |
有两种操作，一是求两点间距离，二是求一点到另一点路径上的第k个点。
LCA妥妥的。
/*by SilverN*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
using namespace std;
const int mxn=;
int read(){
int x=,f=;char ch=getchar();
while(ch<'' || ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>='' && ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
struct edge{
int v,nxt,dis;
}e[mxn<<];
int hd[mxn],mct=;
void add_edge(int u,int v,int d){
e[++mct].v=v;e[mct].nxt=hd[u];e[mct].dis=d;hd[u]=mct;return;
}
int T,n;
int fa[mxn][];
int dep[mxn];
int dis[mxn];
void init(){memset(hd,,sizeof hd);memset(fa,,sizeof fa);mct=;}
void DFS(int u,int f){
dep[u]=dep[f]+;
for(int i=;i<;i++)fa[u][i]=fa[fa[u][i-]][i-];
for(int i=hd[u];i;i=e[i].nxt){
int v=e[i].v;
if(v==f)continue;
fa[v][]=u;
dis[v]=dis[u]+e[i].dis;
DFS(v,u);
}
return;
}
int LCA(int x,int y){
if(dep[x]<dep[y])swap(x,y);
for(int i=;i>=;i--)
if(dep[fa[x][i]]>=dep[y])x=fa[x][i];
if(x==y)return y;
for(int i=;i>=;i--){
if(fa[x][i]!=fa[y][i])x=fa[x][i],y=fa[y][i];
}
return fa[x][];
}
inline int dist(int x,int y){//求距离
int tmp=LCA(x,y);
return dis[x]+dis[y]-dis[tmp]*;
}
inline int find(int x,int k){//上溯
for(int i=;i>=;i--){
if(k&(<<i))x=fa[x][i];
}
return x;
}
inline int solve(int x,int y,int k){//查询从x到y路径上第k个结点
int tmp=LCA(x,y);
int mid=dep[x]-dep[tmp]+;
if(k==mid)return tmp;
if(k>mid){
int dd=dep[y]-dep[tmp]+;
mid=k-mid+;
k=dd-mid;
return find(y,k);
}
else
return find(x,k-);
}
int main(){
T=read();
int i,j,x,y,d;
while(T--){
init();
n=read();
for(i=;i<n;i++){
x=read();y=read();d=read();
add_edge(x,y,d);
add_edge(y,x,d);
}
int rt=n/+;
dis[rt]=;
DFS(rt,);
char op[];
while(scanf("%s",op) && (op[]!='D' || op[]!='O')){
if(op[]=='K'){
x=read();y=read();d=read();
printf("%d\n",solve(x,y,d));
}
if(op[]=='D'){
x=read();y=read();
printf("%d\n",dist(x,y));
}
}
}
return ;
}