Description
A - Breadth-First Search by Foxpower
Problem Statement
Fox Ciel went to JAG Kingdom by bicycle, but she forgot a place where she parked her bicycle. So she needs to search it from a bicycle-parking area before returning home.
The parking area is formed as a unweighted rooted tree TT with nn vertices, numbered 11 through nn. Each vertex has a space for parking one or more bicycles. Ciel thought that she parked her bicycle near the vertex 11, so she decided to search it from there by the breadth-first search. That is, she searches it at the vertices in the increasing order of their distances from the vertex 11. If multiple vertices have the same distance, she gives priority to the vertices in the order of searching at their parents. If multiple vertices have the same parent, she searches at the vertex with minimum number at first.
Unlike a computer, she can't go to a next vertex by random access. Thus, if she goes to the vertex jj after the vertex ii, she needs to walk the distance between the vertices ii and jj. BFS by fox power perhaps takes a long time, so she asks you to calculate the total moving distance in the worst case starting from the vertex 11.
Input
The input is formatted as follows.
nn
p2p2 p3p3 p4p4 ⋯⋯ pnpn
The first line contains an integer nn (1≤n≤1051≤n≤105), which is the number of vertices on the unweighted rooted tree TT. The second line contains n−1n−1 integers pipi (1≤pi<i1≤pi<i), which are the parent of the vertex ii. The vertex 11 is a root node, so p1p1 does not exist.
Output
Print the total moving distance in the worst case in one line.
Sample Input 1
4
1 1 2
Output for the Sample Input 1
6
Sample Input 2
4
1 1 3
Output for the Sample Input 2
4
Sample Input 3
11
1 1 3 3 2 4 1 3 2 9
Output for the Sample Input 25
25
题意：一有一棵树，直接相连的节点之间的距离均为1，一个人站在根节点，从深度从低到高搜索，从每次只能把同一深度的搜索完了才能向下一层搜索，问至少需要经过多少距离才能把所有节点都访问一遍
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
typedef unsigned long long Ull;
#define MM(a,b) memset(a,b,sizeof(a));
const double eps = 1e-10;
const int inf =0x7f7f7f7f;
const double pi=acos(-1);
const int maxn=100000+10;
int c[maxn][22],deep[maxn],vis[maxn];
vector<vector<int> > G(maxn);
void dfs(int u,int par)
{
c[u][0]=par;
for(int i=1;i<=20;i++) c[u][i]=c[c[u][i-1]][i-1];
for(int i=0;i<G[u].size();i++)
{
int v=G[u][i];
if(v==par) continue;
deep[v]=deep[u]+1;
dfs(v,u);
}
}
int lca(int u,int v)
{
if(deep[u]<deep[v]) swap(u,v);
for(int i=20;i>=0;i--)
if(deep[c[u][i]]>=deep[v])
u=c[u][i];
if(u==v) return u;
for(int i=20;i>=0;i--)
if(c[u][i]^c[v][i])
{
u=c[u][i];
v=c[v][i];
}
return c[u][0];
}
int dis(int u,int v)
{
return deep[u]+deep[v]-2*deep[lca(u,v)];
}
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i=1;i<=n;i++) G[i].clear();
deep[1]=0;
for(int i=2;i<=n;i++)
{
int x;scanf("%d",&x);
G[x].push_back(i);
}
dfs(1,1);
int pre=1;ll ans=0;
queue<int> q;
q.push(1);
while(q.size())
{
int u=q.front();q.pop();
for(int i=0;i<G[u].size();i++)
{
int v=G[u][i];
ans+=dis(pre,v);
pre=v;
q.push(v);
}
}
printf("%lld
",ans);
}
return 0;
}
分析:LCA倍增法的第一题，需要设置一个pre代表先前处于哪个位置，然后对于当前需要访问的节点u，求pre和u之间的距离就好，通过lca来求