1021. Deepest Root (25)-PAT

1021. Deepest Root (25)

时间限制

1500 ms

内存限制

32000 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

推荐指数：※※

来源：http://pat.zju.edu.cn/contests/pat-a-practise/1021

本题直接用dfs就可以了（本以为要加一些剪枝才能通过。。。）

#include<iostream>
#include<string.h>
#include<vector>
using namespace std;
#define N 10001
vector<int> edge[N];
int deepest[N];
int visited[N];
int n;
int  dfs(int now_node){
	if(visited[now_node]==1)
		return 0;
	visited[now_node]=1;
	if(edge[now_node].size()==0)
		return 1;
	int i,maxdeep=0,tmp;
	for(i=0;i<edge[now_node].size();i++){//find a deepest from this node
		if(visited[edge[now_node][i]]==0){
			tmp=dfs(edge[now_node][i]);
			maxdeep=maxdeep>tmp?maxdeep:tmp;
		}
	}
	return maxdeep+1;
}
int main()
{
	int i,j;
	cin>>n;
	for(i=1;i<=n-1;i++){
		int v1,v2;
		cin>>v1>>v2;
		edge[v1].push_back(v2);
		edge[v2].push_back(v1);
	}
	memset(deepest,0,sizeof(deepest));
	int max_deep=0;
	int flag=0;
	for(i=1;i<=n;i++){
		if(0==flag)
			memset(visited,0,sizeof(visited));
		deepest[i]=dfs(i);
		max_deep=max_deep>deepest[i]?max_deep:deepest[i];
		for(j=i;j<=n;j++){//check if it is a connected graph
			if(visited[j]==0){
				flag++;
				i=j-1;
				break;
			}
		}
	}
	if(flag>0){
		cout<<"Error: "<<flag+1<<" components"<<endl;
	}
	else{
		for(i=1;i<=n;i++){
			if(deepest[i]==max_deep)
				cout<<i<<endl;
		}
	}
	return 0;
}