POJ 1330 Nearest Common Ancestors(lca)
POJ 1330 Nearest Common Ancestors
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
Output
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
一道关于LCA的算法题,太弱的我不会做。贴出大神的AC代码,仅供参考学习
#include<iostream>
#include<vector>
#define MAX 10010
using namespace std; int n,flag;
int f[MAX],r[MAX],ancestor[MAX];
int indegreen[MAX],vis[MAX];
vector<int> head[MAX],Que[MAX]; void Init()
{
int i,a,b;
cin>>n;
flag=0;
for(i=1;i<=n;i++)
{
head[i].clear();
Que[i].clear();
f[i]=i;
r[i]=1;
ancestor[i]=0;
indegreen[i]=0;
vis[i]=0;
}
for(i=1;i<n;i++)
{
cin>>a>>b;
head[a].push_back(b);
indegreen[b]++;
}
cin>>a>>b;
Que[a].push_back(b);
Que[b].push_back(a);
} int Find(int u)
{
if(f[u]==u)
return f[u];
else
f[u]=Find(f[u]);
return f[u];
} void Union(int v,int u)
{
int a,b;
a=Find(v);
b=Find(u);
if(a==b)
return ;
if(r[a]<=r[b])
{
f[a]=b;
r[b]+=r[a];
}
else
{
f[b]=a;
r[a]+=r[b];
}
} void LCA(int k)
{
int i,size;
size=head[k].size();
ancestor[k]=k;
for(i=0;i<size;i++)
{
if(flag)
break;
LCA(head[k][i]);
Union(k,head[k][i]);
ancestor[Find(k)]=k;
}
vis[k]=1;
size=Que[k].size();
for(i=0;i<size;i++)
{
if(vis[Que[k][i]])
{
flag=1;
cout<<ancestor[Find(Que[k][i])]<<endl;
return ;
}
}
} int main()
{
int T;
cin>>T;
while(T--)
{
Init();
for(int i=1;i<=n;i++)
{
if(!indegreen[i])
{
LCA(i);
break;
}
}
}
return 0;
}
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