POJ - 1330
Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below:
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 Sample Output 4 Source |
这是一道裸LCA，给你一个有根树，再给你两个点判断其最近公共祖先，可以用tarjan解决
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#define X first
#define Y second
using namespace std;
typedef pair<int,int> pii;
const int maxn=;
int f[maxn],n,LCA[maxn],in[maxn],vis[maxn],R;
vector<int> V[maxn];
pii P;
void init()
{
for (int i=; i<=n; i++)
V[i].clear(),f[i]=i;
memset(LCA,,sizeof(LCA));
memset(vis,,sizeof(vis));
memset(in,,sizeof(in));
}
int find(int x)
{
return f[x]==x?x:f[x]=find(f[x]);
}
int mix(int x,int y)
{
int fx=find(x),fy=find(y);
if (fx==fy) return ;
f[fx]=fy;
return ;
}
void Tarjan(int root)
{
vis[root]=;
if (P.X==root&&vis[P.Y])
{
LCA[R]=find(P.Y);
return ;//因为只有一条边，找到直接return
}
if (P.Y==root&&vis[P.X])
{
LCA[R]=find(P.X);
return ;
}
for (int i=; i<V[root].size(); i++)
{
if (!vis[V[root][i]]);
Tarjan(V[root][i]);
f[V[root][i]]=root;
}
}
int main()
{
int T;
scanf("%d",&T);
while (T--)
{
int a,b;
scanf("%d",&n);
init();
for (int i=; i<n; i++)
{
scanf("%d%d",&a,&b);
if (a!=b)
{
in[b]++;//in记录入度
V[a].push_back(b);
}
}
scanf("%d%d",&a,&b);
P.X=a,P.Y=b;
for (int i=;i<=n;i++)
if (in[i]==)//根节点的入度为0
{
R=i;//R为根节点
Tarjan(i);
printf("%d\n",LCA[R]);
break;
}
}
return ;
}