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:

51 21 31 42 5

Sample Output 1:

345

Sample Input 2:

51 31 42 53 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;}