PAT 甲级 1021 Deepest Root (并查集，树的遍历)

1021. Deepest Root (25)

时间限制

1500 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

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 you are supposed to find the root that results in a highest tree. Such a root is called the deepest root.

Input Specification:

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

Output Specification:

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

先求连通块，通过并查集，

然后枚举每一个点dfs，

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <vector>

using namespace std;
const int maxn=1e4;
int n;
struct Node
{
  int value;
  int next;
}edge[maxn*2+5];
int father[maxn+5];
int head[maxn+5];
int vis[maxn+5];
int num[maxn+5];
int tag[maxn+5];
int tot,cnt;
void add(int x,int y)
{
  edge[tot].value=y;
  edge[tot].next=head[x];
  head[x]=tot++;
}
int find(int x)
{
  if(father[x]!=x)
    father[x]=find(father[x]);
  return father[x];
}
void dfs(int root,int deep)
{
  vis[root]=1;
  int tag=0;
  for(int i=head[root];i!=-1;i=edge[i].next)
  {
    int y=edge[i].value;
        if(!vis[y])
    {
      tag=1;
      dfs(y,deep+1);
    }
  }
  if(!tag)
    num[cnt]=max(num[cnt],deep);
}
int main()
{
  scanf("%d",&n);
  int x,y;
  memset(head,-1,sizeof(head));
  for(int i=1;i<=n;i++)
    father[i]=i;
  tot=0;
  for(int i=1;i<n;i++)
  {
        scanf("%d%d",&x,&y);
        int fx=find(x);
    int fy=find(y);
    if(fx!=fy)
      father[fx]=fy;
    add(x,y);
    add(y,x);
  }
  memset(tag,0,sizeof(tag));
  int res=0;
  for(int i=1;i<=n;i++)
  {
    find(i);
    tag[father[i]]=1;
    }
  for(int i=1;i<=n;i++)
       if(tag[i])
       res++;
  if(res>1)
    printf("Error: %d components\n",res);
  else
  {
    for(int i=1;i<=n;i++)
    {
            memset(vis,0,sizeof(vis));
      cnt=i;
      dfs(i,0);
    }
    int ans=0;
        for(int i=1;i<=cnt;i++)
      ans=max(ans,num[i]);
    for(int i=1;i<=cnt;i++)
      if(num[i]==ans)
        printf("%d\n",i);
  }
  return 0;
}