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

并查集的相关资料前面已经有介绍了，所以用并查集判断是否是树很简单，然后就是最长根的查找，这里用的是dfs，其实bfs更容易解该问题。

查找最长根的时候，首先从一点进行深搜，把其中距离最远的点记录下来（可访问到所有顶点，因为已经保证是树），然后从这些最远点中任选一点再进行深度搜索，这个时候的所有最远点都是最长根，关键是第一遍dfs保存的最远点也是最长根，这个地方最难理解了，首先，若第一遍dfs后的最远根不在从1开始的同一条支路上（他们的最近公共祖先是1），这是这些点必定都会是将来的最长根这很明显，若第一遍dfs后的最远根在从1开始的同一条支路上(他们的最近公共祖先不是1记为f)，则第二遍dfs时的最长根所在的dfs遍历序列必过f，所以此时的情况又可将f点看做第一种情况中的1节点类似。综上第一次与第二次dfs的合集才是最终结果。

#include <iostream>
#include <cstdio>
#include <map>
#include <vector>
#include <set>using namespace std;const int maxx = 10001;map<int,vector<int> > tree;
set<int> res;
set<int> tres;
int pre[maxx],cnt,maxn = -1;
int visit[maxx];void dfs(int p,int step){if(step>maxn){res.clear();res.insert(p);maxn = step;}else if(step==maxn){res.insert(p);}vector<int>::iterator ite = tree[p].begin();for(;ite!=tree[p].end();++ite){if(visit[*ite]!=1){visit[*ite] = 1;dfs(*ite,step+1);visit[*ite] = 0;}}
}void init(int n){int i;for(i=1;i<=n;++i){pre[i]=i;visit[i]=0;}
}int root(int x){if(x!=pre[x]){pre[x] = root(pre[x]);}return pre[x];
}void merge(int x,int y){int fa = root(x);int fb = root(y);if(fa!=fb){pre[fa] = fb;--cnt;}
}int main(){int n,i,a,b;set<int>::iterator ite;scanf("%d",&n);cnt = n-1;init(n);for(i=1;i<n;++i){scanf("%d %d",&a,&b);tree[a].push_back(b);tree[b].push_back(a);merge(a,b);}if(cnt!=0){printf("Error: %d components\n",1+cnt);return 0;}visit[1] = 1;dfs(1,1);visit[1] = 0;//因为已经是树，所以不会再有环了，所以所有dfs后都恢复该点的可访问性很重要ite = res.begin();for(;ite!=res.end();++ite){tres.insert(*ite);}int point = (*res.begin());visit[point] = 1;dfs(point,1);visit[point] = 0;ite = res.begin();for(;ite!=res.end();++ite){tres.insert(*ite);}ite = tres.begin();for(;ite!=tres.end();++ite){printf("%d\n",*ite);}return 0;
}

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