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POJ 1330 Nearest Common Ancestors
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Nearest Common Ancestors
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 26612 | Accepted: 13734 |
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
The input consists of T test cases. The number of test 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
Print exactly one line for each test case. The line 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
思路
最近公共祖先模板题
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
const int maxn = 10005;
struct Edge{
int to,next;
}edge[maxn];
vector<int>qry[maxn];
int N,tot,fa[maxn],head[maxn],indegree[maxn],ancestor[maxn];
bool vis[maxn];
void init()
{
tot = 0;
for (int i = 1;i <= N;i++) fa[i] = i,head[i] = -1,indegree[i] = 0,vis[i] = false,qry[i].clear();
}
void addedge(int u,int to)
{
edge[tot] = (Edge){to,head[u]};
head[u] = tot++;
}
int find(int x)
{
int r = x;
while (r != fa[r]) r = fa[r];
int i = x,j;
while (i != r)
{
j = fa[i];
fa[i] = r;
i = j;
}
return r;
}
void Union(int x,int y)
{
x = find(x),y = find(y);
if (x == y) return;
fa[y] = x; //不能写成fa[x] = y,与集合合并的祖先有关系
}
void targin_LCA(int u)
{
ancestor[u] = u;
for (int i = head[u];i != -1;i = edge[i].next)
{
int v = edge[i].to;
targin_LCA(v);
Union(u,v);
ancestor[find(u)] = u;
}
vis[u] = true;
int size = qry[u].size();
for (int i = 0;i < size;i++)
{
if (vis[qry[u][i]]) printf("%d\n",ancestor[find(qry[u][i])]);
return;
}
}
int main()
{
int T;
scanf("%d",&T);
while (T--)
{
int u,v;
scanf("%d",&N);
init();
for (int i = 1;i < N;i++)
{
scanf("%d%d",&u,&v);
addedge(u,v);
indegree[v]++;
}
scanf("%d%d",&u,&v);
qry[u].push_back(v),qry[v].push_back(u);
for (int i = 1;i <= N;i++)
{
if (!indegree[i])
{
targin_LCA(i);
break;
}
}
}
return 0;
}
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<vector>
using namespace std;
const int MAXN=10010;
int F[MAXN];//并查集
int r[MAXN];//并查集中集合的个数
bool vis[MAXN];//访问标记
int ancestor[MAXN];//祖先
struct Node
{
int to,next;
}edge[MAXN*2];
int head[MAXN];
int tol;
void addedge(int a,int b)
{
edge[tol].to=b;
edge[tol].next=head[a];
head[a]=tol++;
edge[tol].to=a;
edge[tol].next=head[b];
head[b]=tol++;
}
struct Query
{
int q,next;
int index;//查询编号
}query[MAXN*2];//查询数
int answer[MAXN*2];//查询结果
int cnt;
int h[MAXN];
int tt;
int Q;//查询个数
void add_query(int a,int b,int i)
{
query[tt].q=b;
query[tt].next=h[a];
query[tt].index=i;
h[a]=tt++;
query[tt].q=a;
query[tt].next=h[b];
query[tt].index=i;
h[b]=tt++;
}
void init(int n)
{
for(int i=1;i<=n;i++)
{
F[i]=-1;
r[i]=1;
vis[i]=false;
ancestor[i]=0;
tol=0;
tt=0;
cnt=0;//已经查询到的个数
}
memset(head,-1,sizeof(head));
memset(h,-1,sizeof(h));
}
int find(int x)
{
if(F[x]==-1)return x;
return F[x]=find(F[x]);
}
void Union(int x,int y)//合并
{
int t1=find(x);
int t2=find(y);
if(t1!=t2)
{
if(r[t1]<=r[t2])
{
F[t1]=t2;
r[t2]+=r[t1];
}
else
{
F[t2]=t1;
r[t1]+=r[t2];
}
}
}
void LCA(int u)
{
//if(cnt>=Q)return;//不要加这个
ancestor[u]=u;
vis[u]=true;//这个一定要放在前面
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(vis[v])continue;
LCA(v);
Union(u,v);
ancestor[find(u)]=u;
}
for(int i=h[u];i!=-1;i=query[i].next)
{
int v=query[i].q;
if(vis[v])
{
answer[query[i].index]=ancestor[find(v)];
cnt++;//已经找到的答案数
}
}
}
bool flag[MAXN];
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T;
int N;
int u,v;
scanf("%d",&T);
while(T--)
{
scanf("%d",&N);
init(N);
memset(flag,false,sizeof(flag));
for(int i=1;i<N;i++)
{
scanf("%d%d",&u,&v);
flag[v]=true;
addedge(u,v);
}
Q=1;//查询只有一组
scanf("%d%d",&u,&v);
add_query(u,v,0);//增加一组查询
int root;
for(int i=1;i<=N;i++)
if(!flag[i])
{
root=i;
break;
}
LCA(root);
for(int i=0;i<Q;i++)//输出所有答案
printf("%d\n",answer[i]);
}
return 0;
}
POJ 1330 Nearest Common Ancestors
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