posted on 2019-08-28 16:45:21
A. tree traversal
Title Description
Given a \ (n-\) unrooted tree nodes (node number \ (0 \) to \ (. 1-n-\) ) and a node \ (X \) , please \ (X \) No. Node is the root, do a DFS with a BFS .
Input Format
Data is read from the standard input.
The first line enter a positive integer \ (n-\) ( \ (. 1 \ n-Leq \ Leq 200000 \) ), representative of the number of nodes of this tree.
$ N-1 $ next row (row number from $ 1 to $ $ n-1 $), the first \ (I \) line enter a positive integer a_i $ \ ((\) 0 \ Leq a_i \ I $ Leq) , represents the \ (I \) attached there is an edge between the node and the second node $ $ a_i.
The last line of input X $ \ ((\) 0 \ Leq X <n-$), representative of the root node number.
Output Format
Output to standard output.
$ 2 $ output lines of $ $ n-number, the first row represents $ 1 $ DFS sequence, the second row represents $ 2 $ BFS order.
Note: If a node has multiple son, then the son should be in the order of decreasing number to traverse.
Sample input 1
7
0
1
0
0
1
4
1Sample output
1 5 2 0 4 6 3
1 5 2 0 4 3 6 solution:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <vector>
using namespace std;
vector<int> to[200005];
int n,a,vis[200005],q[200005],start;
int cmp(int b,int c)
{
return b>c;
}
void ad(int u,int v)
{
to[u].push_back(v);
}
void dfs(int k)
{
cout << k << ' ';
for(int i=0;i<to[k].size();i++)
{
if(vis[to[k][i]]==0)
{
vis[to[k][i]]=1;
dfs(to[k][i]);
}
}
}
void bfs(int k)
{
int head=1,tail=1;
q[tail]=k;
tail++;
while(head<tail)
{
int now=q[head];
cout << now << " ";
head++;
for(int i=0;i<to[now].size();i++)
{
if(vis[to[now][i]]==0)
{
vis[to[now][i]]=1;
q[tail]=to[now][i];
tail++;
}
}
}
}
int main()
{
cin >> n;
for(int i=1;i<=n-1;i++)
{
cin >> a;
ad(a,i);
ad(i,a);
}
cin >> start;
for(int i=0;i<n;i++)
sort(to[i].begin(),to[i].end(),cmp);
/*for(int i=0;i<n;i++)
{
for(int j=0;j<to[i].size();j++)
cout << to[i][j] << " ";
cout << endl;
}*/
//对于有根树，将start换成根节点即可
vis[start]=1;
dfs(start);
memset(vis,0,sizeof(vis));
cout << endl;
vis[start]=1;
bfs(start);
return 0;
}
B. center and diameter of the tree
Title Description
Given a \ (n-\) unrooted tree nodes (node number \ (0 \) to \ (1-n-\) ), all side lengths are \ (1 \) diameter, determined the tree length.
Definition tree of the center of the tree node with the smallest maximum value of distances from all nodes (the center of a tree may be more than one), the output of the center of the tree.
Input Format
Data is read from the standard input.
The first line enter a positive integer \ (n-\) ( \ (. 1 \ n-Leq \ Leq 200000 \) ), representative of the number of nodes of this tree.
Next \ (n-1 \) (line numbered \ (1 \) to \ (n-1 \) ), the first \ (I \) line enter a positive integer \ (a_i \) ( \ (0 \ a_i Leq <I \) ), represents the \ (I \) node and the second \ (a_i \) attached there is an edge between the nodes.
Output Format
Output to standard output.
Output \ (2 \) th, \ (1 \) line an integer, representative of the diameter of the tree; second \ (2 \) line outputs in accordance with a number of integers in ascending numerical order, node number representative of the center of the tree.
Sample input 1
6
0
1
0
0
1Sample output
3
0 1 solution:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <vector>
using namespace std;
vector<int> to[200005];
int n, a, vis[200005], q[200005], step[200005][2], head = 1, tail = 1, cen[3];
int cmp(int b, int c) { return b > c; }
void ad(int u, int v) { to[u].push_back(v); }
void bfs(int k) {
head = 1;
tail = 1;
q[tail] = k;
tail++;
while (head < tail) {
int now = q[head];
head++;
for (int i = 0; i < to[now].size(); i++) {
if (vis[to[now][i]] == 0) {
vis[to[now][i]] = 1;
q[tail] = to[now][i];
step[tail][0] = step[head - 1][0] + 1;
step[tail][1] = head - 1;
tail++;
}
}
}
}
int main() {
cin >> n;
for (int i = 1; i <= n - 1; i++) {
cin >> a;
ad(a, i);
ad(i, a);
}
for (int i = 0; i < n; i++) sort(to[i].begin(), to[i].end(), cmp);
vis[0] = 1;
bfs(0);
int point = q[head - 1];
memset(vis, 0, sizeof(vis));
memset(step, 0, sizeof(step));
vis[point] = 1;
bfs(point);
int lenth = step[tail - 1][0];
cout << lenth << endl;
if (lenth % 2 == 0) {
for (int i = tail - 1; i > 0; i = step[i][1]) {
if (step[i][0] == lenth / 2) {
cen[1] = q[i];
cout << cen[1] << endl;
return 0;
}
}
} else {
for (int i = tail - 1; i > 0; i = step[i][1]) {
if (step[i][0] == (lenth + 1) / 2) {
cen[1] = q[i];
}
if (step[i][0] == (lenth - 1) / 2) {
cen[2] = q[i];
if (cen[2] < cen[1])
swap(cen[2], cen[1]);
cout << cen[1] << " " << cen[2] << endl;
return 0;
}
}
}
}