洛谷 SP14932 LCA - Lowest Common Ancestor

洛谷 SP14932 LCA - Lowest Common Ancestor

洛谷评测传送门

题目描述

A tree is an undirected graph in which any two vertices are connected by exactly one simple path. In other words, any connected graph without cycles is a tree. - Wikipedia

The lowest common ancestor (LCA) is a concept in graph theory and computer science. Let T be a rooted tree with N nodes. The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself). - Wikipedia

Your task in this problem is to find the LCA of any two given nodes v and w in a given tree T.

For example the LCA of nodes 9 and 12 in this tree is the node number 3.

Input

The first line of input will be the number of test cases. Each test case will start with a number N the number of nodes in the tree, 1 <= N <= 1,000. Nodes are numbered from 1 to N. The next N lines each one will start with a number M the number of child nodes of the Nth node, 0 <= M <= 999 followed by M numbers the child nodes of the Nth node. The next line will be a number Q the number of queries you have to answer for the given tree T, 1 <= Q <= 1000. The next Q lines each one will have two number v and w in which you have to find the LCA of v and w in T, 1 <= v, w <= 1,000.

Input will guarantee that there is only one root and no cycles.

Output

For each test case print Q + 1 lines, The first line will have “Case C:” without quotes where C is the case number starting with 1. The next Q lines should be the LCA of the given v and w respectively.

Example

Input:
1
7
3 2 3 4
0
3 5 6 7
0
0
0
0
2
5 7
2 7

Output:
Case 1:
3
1

输入格式

无

输出格式

无

题意翻译

Description:

一棵树是一个简单无向图，图中任意两个节点仅被一条边连接，所有连通无环无向图都是一棵树。-Wikipedia

最近公共祖先（LCA）是……（此处省去对LCA的描述），你的任务是对一棵给定的树TT以及上面的两个节点u,vu,v求出他们的LCALCA。

例如图中99和1212号节点的LCA*L*C*A*为33号节点

Input:

输入的第一行为数据组数TT，对于每组数据，第一行为一个整数N(1\leq N\leq1000)N(1≤N≤1000)，节点编号从11到NN，接下来的NN行里每一行开头有一个数字M(0\leq M\leq999)M(0≤M≤999)，MM为第ii个节点的子节点数量，接下来有MM个数表示第ii个节点的子节点编号。下面一行会有一个整数Q(1\leq Q\leq1000)Q(1≤Q≤1000)，接下来的QQ行每行有两个数u,vu,v，输出节点u,vu,v在给定树中的LCALCA。

输入数据保证只有一个根节点并且没有环。

Output:

对于每一组数据输出Q+1Q+1行，第一行格式为"Case i:"（没有双引号），i表示当前数据是第几组，接下来的QQ行每一行一个整数表示一对节点u,vu,v的LCALCA。

Sample Input:

1
7
3 2 3 4
0
3 5 6 7
0
0
0
0
2
5 7
2 7

Sample Output:

Case 1:
3
1

Translated by @yxl_gl

输入输出样例

无

题解：

LCA模板题目双倍经验~~

点进来的小伙伴肯定还不太会LCA...

请参考蒟蒻的这篇博客：

（这里介绍了倍增求LCA，其实求LCA还有好多方式，比如离线Tarjan和树链剖分等，有兴趣的巨佬可以自己涉及，如果只求LCA的话，还是这种倍增法更快一些）

求解LCA问题的几种方式

当然，本题还有一些小细节，比如多组数据数据要清空，以及比较奇葩的读入边的方式。

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1010;
char *p1,*p2,buf[100000];
#define nc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++)
int read()
{
	int x=0,f=1;
	char ch=nc();
	while(ch<48){if(ch=='-')f=-1;ch=nc();}
	while(ch>47)	x=(((x<<2)+x)<<1)+ch-48,ch=nc();
	return x*f;
}
int n,m,q;
int tot,head[maxn],nxt[maxn<<1],to[maxn<<1];
int deep[maxn],fa[maxn][21];
void add(int x,int y)
{
	to[++tot]=y;
	nxt[tot]=head[x];
	head[x]=tot;
}
void dfs(int x,int f)
{
	deep[x]=deep[f]+1;
	fa[x][0]=f;
	for(int i=1;(1<<i)<=deep[x];i++)
		fa[x][i]=fa[fa[x][i-1]][i-1];
	for(int i=head[x];i;i=nxt[i])
	{
		int y=to[i];
		if(y==f)
			continue;
		dfs(y,x);
	}
}
int lca(int x,int y)
{
	int ret;
	if(deep[x]<deep[y])
		swap(x,y);
	for(int i=20;i>=0;i--)
		if(deep[fa[x][i]]>=deep[y])
			x=fa[x][i];
	if(x==y)
		return y;
	for(int i=20;i>=0;i--)
	{
		if(fa[x][i]!=fa[y][i])
		{
			x=fa[x][i];
			y=fa[y][i];
		}
		else
			ret=fa[x][i];
	}
	return ret;
}
int main()
{
	int t;
	t=read();
	for(int k=1;k<=t;k++)
	{
		tot=0;
		memset(head,0,sizeof(head));
		memset(nxt,0,sizeof(nxt));
		memset(to,0,sizeof(to));
		memset(deep,0,sizeof(deep));
		memset(fa,0,sizeof(fa));
		n=read();
		for(int i=1;i<=n;i++)
		{
			m=read();
			if(!m)
				continue;
			for(int j=1;j<=m;j++)
			{
				int u=read();
				add(u,i);
				add(i,u);
			}
		}
		dfs(1,0);
		q=read();
		printf("Case %d:\n",k);
		while(q--)
		{
			int u=read();
			int v=read();
			printf("%d\n",lca(u,v));
		}
	}
	return 0;
}