1021. Deepest Root (25)-PAT

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本文介绍了一种使用深度优先搜索算法解决寻找图中最高树的最深根节点的方法，并提供了完整的C++代码实现。该算法适用于连接且无环的图（即树），能够找出使得树高度最大的根节点。

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:

Sample Output 1:

3
4
5

Sample Input 2:

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;
}