1021. Deepest Root (25) and check set &&dfs

1021. Deepest Root (25)
Time limit
Ms
Memory limit
65536 KB
Code Length Limitations
16000 B
Procedures for the award of questions
Standard
Author
CHEN, Yue

A graph which is connected and acyclic can be considered a tree. The height of the tree depends on the selected root. Now is supposed to find the root of the results in a highest tree. Such A root is called the deepest root.

Input Specification:

Each input file contains the one test case. The first line contains a positive an integer N (<=10000) which is the number of nodes, and hence the nodes is numbered from 1 to N. Then N-1 lines follow, each describes a edge by given the adjacent nodes ' numbers.

Output Specification:

For each test case, print each of the deepest roots in a line. If such a root is isn't unique, print them in increasing order of their numbers. In case that the given graph isn't a tree, print "Error:k components" where K is the number of connected He graph.
Sample Input 1:

5
1 2
1 3
1 4
2 5

Sample Output 1:

3
4
5

Sample Input 2:

5
1 3
1 4
2 5
3 4

Sample Output 2:

Error:2 components

Submit Code

It's important to note that the graph is stored with adjacency tables if the adjacency matrix is timed out (sparse graph)

The first time a DFS is written with a return value, it is not very skilled

#include <iostream> #include <cstdio> #include <cstring> #include <vector> #include < algorithm>using namespace Std;const int maxn=10001;bool vis[maxn];int a[maxn];int n;int cnt=0;vector<int> ma[    maxn];//adjacency Table int d[maxn];//calculates the maximum depth int dfs (int s) When this point is the root, and//dfs represents the maximum depth {int ans=0, which can be reached starting with s;    if (Vis[s]) return 0;//this point has been visited, it is apparent that the maximum depth of this point is 0 vis[s]=true;    int m=ma[s].size (); for (int i=0; i<m; i++) {if (!vis[ma[s][i]]) {int Tmp=dfs (ma[s][i]);//The maximum depth that can be reached with the current point as a starting point, also It is to find the maximum Ans=max (ans,tmp) that can be reached in the adjacency point of S, and the maximum depth of the//ans record is {}} return ans+1;//can go here to show that the depth of s can be added one, that is, S. The depth of the adjacent vertex, the maximum value, plus the S point, so is ans+1}void init (int n)//This is the initialization of the set {for (int i=0; i<=n; i++) a[i]=i;}    int find (int x)//And find the father of X, with path compression {if (a[x]!=x) A[x]=find (a[x]); return a[x];}    void Unio (int x,int y)//merge x, Y to together {x=find (x);    Y=find (y);    if (A[x]==a[y]) return; A[x]=y;}    int main () {int i,j,k,t; Freopen ("In.txt", "R", stdin);   cin>>n;    Init (n);        for (I=1; i<n; i++) {int s,e;        cin>>s>>e;        Unio (s,e);        Ma[s].push_back (e);    Ma[e].push_back (s);    } int sum=0;//determines the number of connected components for (i=1; i<=n; i++) {if (a[i]==i) sum++;        } if (sum>1) {printf ("Error:%d components\n", sum);    return 0;                } else for (i=1; i<=n; i++) {memset (vis,0,sizeof (VIS));            D[i]=dfs (i);    } int Maxv=-1;int index=0;    for (i=1;i<=n;i++) if (D[I]&GT;MAXV) {maxv=d[i];index=i;} for (j=1;j<=n;j++) if (D[j]==d[index]) printf ("%d\n", j);}

1021. Deepest Root (25) and check set &&dfs