POJ题目1330 Nearest Common Ancestors（LCA）

标签：lca

Description

A rooted tree is a well-known data structure in computer science and engineering. An example is shown below:


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

Source

Taejon 2002

ac代码

#include<stdio.h>
#include<string.h>
struct s
{
    int u,v,next;
}edge[10010];
int vis[10010],head[10010],a,b,pre[10010],dig[10010],cnt,ans,flag,n;
void init()
{
    memset(vis,0,sizeof(vis));
    memset(head,-1,sizeof(head));
    memset(dig,0,sizeof(dig));
    cnt=0;
    flag=0;
}
void add(int u,int v)
{
    edge[cnt].u=u;
    edge[cnt].v=v;
    edge[cnt].next=head[u];
    head[u]=cnt++;
}
int find(int x)
{
    if(x==pre[x])
        return x;
    return find(pre[x]);
}
void tarjan(int u)
{
    pre[u]=u;
    vis[u]=1;
    if(flag)
        return;
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(!vis[v])
        {
            tarjan(v);
            pre[v]=u;
        }
    }
    if(u==a&&vis[b])
    {
        ans=find(b);
        flag=1;
    }
    else
    {
        if(u==b&&vis[a])
        {
            ans=find(a);
            flag=1;
        }
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int i,j;
        init();
        scanf("%d",&n);
        for(i=0;i<n-1;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            add(u,v);
            dig[v]++;
        }
        scanf("%d%d",&a,&b);
        for(i=1;i<=n;i++)
        {
            if(dig[i]==0)
            {
                tarjan(i);
                break;
            }
        }
        printf("%d\n",ans);
    }
}

POJ题目1330 Nearest Common Ancestors（LCA）

标签：lca

原文地址：http://blog.csdn.net/yu_ch_sh/article/details/45080997