1021. Deepest Root (25)

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

题目大意：如果是连通图，则求连通图中点Vi到点Vj所有路径中最长(包含多对)的并打印出所有Vi与Vj。如果是非连通图，则打印出有几个子图。用并查集判断是否是连通图，随后用搜索来求最长路。

#include<iostream>
#include<stdio.h>
#include<vector>
#include<cstring>
#include<queue>
using namespace std;
#define max 10002
int visit[max];
int a[max];
int N;
int distan[max];
vector<int>map[max];
int find(int x){
	if(a[x]== x)return x;
	find(a[x]);
}
void unio(int x,int y){
	x = find(x);
	y = find(y);
	if(x != y){
		a[x] = y;
	}
}
int DFS(int key){
	if(visit[key]==1)return 0;
	visit[key]=1;
	int i;
	int sum = 0;
	int m = map[key].size();
	for(i=0;i<m;i++){
		if(visit[map[key][i]]==0){
			int tmp = DFS(map[key][i]);
			if(sum < tmp){
				sum = tmp;
			}
		}
	}
	return sum+1;
}
int main(){
	scanf("%d",&N);
	int i,j,t;
	int s,d;
	for(i=1;i<=N;i++){
		a[i]=i;
	}
	for(i=1;i<N;i++){
		scanf("%d%d",&s,&d);
		unio(s,d);
		map[s].push_back(d);
		map[d].push_back(s);
	}
	int flag=0;
	for(i=1;i<=N;i++){
		if(a[i]==i){
			flag++;
		}
	}
	if(flag>1){
		printf("Error: %d components",flag);
	} else{
		for(i=1;i<=N;i++){
			memset(visit,0,sizeof(visit));
			distan[i]=DFS(i);
		}
		int a=-1,b=0;
		for(i=1;i<=N;i++){
			if(distan[i]>a){
				a=distan[i];
				b=i;
			}
		}
		for(i=1;i<=N;i++){
			if(distan[i] == distan[b]){
				printf("%d\n",i);
			}
		}
	}
	return 0;
}