题目背景
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题目描述
Bessie and Jonell are great friends. Since Farmer John scrambles where the cows graze every day, they are sometimes quite far from each other and can‘t talk.
The pastures and paths on FJ‘s farm form a ‘tree‘ structure. Each pasture has exactly one distinct path to any other pasture, and each pasture (except pasture #1, the ‘root‘) also has a single parent node.
Bessie and Jonell have decided that they will always meet at the closest pasture that that is both an ancestor of Jonell‘s pasture and of Bessie‘s pasture.
FJ created a map of his N (1 <= N <= 1,000) pastures (conveniently numbered 1..N) that tells the parent P_i (1 <= P_i <= N) of each pasture except pasture 1, which has no parent.
FJ has released his daily grazing schedule for the next M (1 <= M <= 1,000) days, so Bessie and Jonell are deciding where they should meet each day for gossip. On day k, Bessie is in pasture B_k (1 <= B_k <= N) and Jonell is in pasture J_k (1 <= J_k <= N).
Given a map and schedule, help Bessie and Jonell find their meeting places.
Consider, for example, the following farm layout:
Pasture Parent Pasture
[1] --------- ----------------
/ | \ 1 ---
/ | \ 2 1
[2] [3] [6] 3 1
/ | \ 4 2
/ | \ 5 8
[4] [8] [9] 6 1
/ \ 7 8
/ \ 8 6
[5] [7] 9 6
Here are the meeting places that Bessie and Jonell would choose
given a six day schedule of their initial grazing locations:
Bessie Jonell Meeting Place
-------- -------- ---------------
2 7 1
4 2 2
1 1 1
4 1 1
7 5 8
9 5 6输入输出格式
输入格式：
- Line 1: Two space-separated integers: N and M
- Lines 2..N: Line i contains a single integer that describes the parent of pasture i: P_i
- Lines N+1..N+M: Line k+N describes Bessie and Jonell‘s respective pastures with two space-separated integers: B_k and J_k
输出格式：
- Lines 1..M: Line j contains the meeting place Bessie and Jonell would use for line j+N of the input
输入输出样例
输入样例#1：
9 6 1 1 2 8 1 8 6 6 2 7 4 2 3 3 4 1 7 5 9 5
输出样例#1：
1 2 3 1 8 6
lca模板题
#include <vector>
#include <cstdio>
#define N 1005
using std::vector;
int n,m,dad[N][25],dep[N],siz[N];
vector<int>G[N];
void dfs(int x)
{
dep[x]=dep[dad[x][0]]+1;
for(int i=0;dad[x][i];++i)
dad[x][i+1]=dad[dad[x][i]][i];
for(int i=0;i<G[x].size();++i)
{
int v=G[x][i];
if(dad[x][0]!=v)
{
dad[v][0]=x;
dfs(v);
}
}
}
inline void swap(int &m,int &n)
{
int tmp=n;
n=m;
m=tmp;
}
int lca(int x,int y)
{
if(dep[x]>dep[y]) swap(x,y);
for(int i=20;i>=0;--i)
if(dep[dad[y][i]]>=dep[x]) y=dad[y][i];
if(x==y) return x;
for(int i=20;i>=0;--i)
if(dad[x][i]!=dad[y][i]) x=dad[x][i],y=dad[y][i];
return dad[x][0];
}
int main()
{
scanf("%d%d",&n,&m);
for(int x,i=2;i<=n;++i)
{
scanf("%d",&x);
G[x].push_back(i);
G[i].push_back(x);
}
dfs(1);
for(int x,y;m--;)
{
scanf("%d%d",&x,&y);
printf("%d\n",lca(x,y));
}
return 0;
}