| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 24762 | Accepted: 12868 |
Description
In the figure, each node is labeled with an integer
from {1, 2,...,16}. Node 8 is the root of the tree. Node x is an ancestor of
node y if node x is in the path between the root and node y. For example, node 4
is an ancestor of node 16. Node 10 is also an ancestor of node 16. As a matter
of fact, nodes 8, 4, 10, and 16 are the ancestors of node 16. Remember that a
node is an ancestor of itself. Nodes 8, 4, 6, and 7 are the ancestors of node 7.
A node x is called a common ancestor of two different nodes y and z if node x is
an ancestor of node y and an ancestor of node z. Thus, nodes 8 and 4 are the
common ancestors of nodes 16 and 7. A node x is called the nearest common
ancestor of nodes y and z if x is a common ancestor of y and z and nearest to y
and z among their common ancestors. Hence, the nearest common ancestor of nodes
16 and 7 is node 4. Node 4 is nearer to nodes 16 and 7 than node 8 is.
For other examples, the nearest common ancestor of nodes 2 and 3 is node
10, the nearest common ancestor of nodes 6 and 13 is node 8, and the nearest
common ancestor of nodes 4 and 12 is node 4. In the last example, if y is an
ancestor of z, then the nearest common ancestor of y and z is y.
Write a
program that finds the nearest common ancestor of two distinct nodes in a tree.
Input
cases (T) is given in the first line of the input file. Each test case starts
with a line containing an integer N , the number of nodes in a tree,
2<=N<=10,000. The nodes are labeled with integers 1, 2,..., N. Each of the
next N -1 lines contains a pair of integers that represent an edge --the first
integer is the parent node of the second integer. Note that a tree with N nodes
has exactly N - 1 edges. The last line of each test case contains two distinct
integers whose nearest common ancestor is to be computed.
Output
should contain the integer that is the nearest common ancestor.
Sample Input
2
16
1 14
8 5
10 16
5 9
4 6
8 4
4 10
1 13
6 15
10 11
6 7
10 2
16 3
8 1
16 12
16 7
5
2 3
3 4
3 1
1 5
3 5
Sample Output
4
3
Source
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
#define N 10010
int n,t,cx,cy,deep[N];
vector<int>p[N<<];
int g[N][],vis[N];
void dfs(int x,int de){
for(int i=;i<p[x].size();i++){
if(!deep[p[x][i]]){
deep[p[x][i]]=deep[x]+;
g[p[x][i]][]=x;
dfs(p[x][i],de+);
}
}
}
int lca(int a,int b){
if(deep[a]<deep[b]) swap(a,b);
int t=deep[a]-deep[b];
for(int i=;i<=;i++){
if((<<i)&t){
a=g[a][i];
}
}
if(a==b) return a;
for(int i=;i>=;i--){
if(g[a][i]!=g[b][i]){
a=g[a][i];
b=g[b][i];
}
}
return g[a][];
}
int main(){
scanf("%d",&t);
while(t--){
memset(g,,sizeof g);
memset(p,,sizeof p);
memset(vis,,sizeof vis);
memset(deep,,sizeof deep);
scanf("%d",&n);
for(int i=,x,y;i<n;i++){
scanf("%d%d",&x,&y);
p[x].push_back(y);
vis[y]++;
}
scanf("%d%d",&cx,&cy);
for(int i=;i<=n;i++){
if(!vis[i]){//注意这是树，所以边是单向的，深搜的时候从根节点开始搜
dfs(i,);
break;
}
} for(int j=;j<=;j++){
for(int i=;i<=n;i++){
g[i][j]=g[g[i][j-]][j-];
}
}
printf("%d\n",lca(cx,cy));
}
return ;
}