LCA - Lowest Common Ancestor
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A tree is an undirected graph in which any two vertices are connected by exactly one simple path. In other words, any connected graph without cycles is a tree. - Wikipedia
The lowest common ancestor (LCA) is a concept in graph theory and computer science. Let T be a rooted tree with N nodes. The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself). - Wikipedia
Your task in this problem is to find the LCA of any two given nodes v and w in a given tree T.
For example the LCA of nodes 9 and 12 in this tree is the node number 3.
Input
The first line of input will be the number of test cases. Each test case will start with a number N the number of nodes in the tree, 1 <= N <= 1,000. Nodes are numbered from 1 to N. The next N lines each one will start with a number M the number of child nodes of the Nth node, 0 <= M <= 999 followed by M numbers the child nodes of the Nth node. The next line will be a number Q the number of queries you have to answer for the given tree T, 1 <= Q <= 1000. The next Q lines each one will have two number v and w in which you have to find the LCA of v and w in T, 1 <= v, w <= 1,000.
Input will guarantee that there is only one root and no cycles.
Output
For each test case print Q + 1 lines, The first line will have “Case C:” without quotes where C is the case number starting with 1. The next Q lines should be the LCA of the given v and w respectively.
Example
Input:1
7
3 2 3 4
0
3 5 6 7
0
0
0
0
2
5 7
2 7
Output:
Case 1: 3 1
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#define INF 1e9
#define MOD 1000000007
#define maxn 1005
using namespace std;
typedef long long ll;
vector<int> v[maxn];
int vis[maxn], d[maxn<<2], p[maxn<<2], c[maxn], cnt;
int dp[maxn<<2][20];
void dfs(int i, int dis, int f){
c[i] = cnt;
d[cnt] = dis;
p[cnt++] = i;
for(int j = 0; j < v[i].size(); j++){
int h = v[i][j];
if(h != f){
dfs(h, dis+1, i);
d[cnt] = dis;
p[cnt++] = i;
}
}
}
void RMQ(){
for(int i = 0; i < cnt; i++)
dp[i][0] = p[i];
for(int i = 1; (1<<i) <= cnt; i++)
for(int j = 0; j+(1<<i) <= cnt; j++){
int k1 = dp[j][i-1], k2 = dp[j+(1<<(i-1))][i-1];
int ans = k2;
if(d[c[k1]] < d[c[k2]])
ans = k1;
dp[j][i] = ans;
}
}
int main(){
// freopen("in.txt", "r", stdin);
int t, cas = 0;
scanf("%d", &t);
while(t--){
cnt = 0;
int n, m, a, b;
scanf("%d", &n);
memset(vis, 0, sizeof(vis));
for(int i = 1; i <= n; i++)
v[i].clear();
for(int i = 1; i <= n; i++){
scanf("%d", &m);
for(int j = 0; j < m; j++){
scanf("%d", &a);
vis[a] = 1;
v[i].push_back(a);
}
}
for(int i = 1; i <= n; i++){
if(vis[i] == 0){
dfs(i, 0, -1);
break;
}
}
RMQ();
int q;
scanf("%d", &q);
printf("Case %d:\n", ++cas);
while(q--){
scanf("%d%d", &a, &b);
int p1 = c[a], p2 = c[b];
if(p1 > p2)
swap(p1, p2);
int r = 0;
while(1<<(r+1) <= p2-p1+1)
r++;
int k1 = dp[p1][r], k2 = dp[p2-(1<<r)+1][r];
if(d[c[k1]] < d[c[k2]])
printf("%d\n", k1);
else
printf("%d\n", k2);
}
}
return 0;
}
