标签：eof input 存在 ace void find ios scribe error
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 5Sample Output 1:
3 4 5Sample Input 2:
5 1 3 1 4 2 5 3 4Sample Output 2:
Error: 2 components
先随便找个点找出离这个点最远的点，首先这些点是符合的，然后从中拿一点做同样操作。
可以这样考虑，最深的根至少存在两个，至少是两端一端一个，中间的点必定离其中一个或多个端点最远所以随便找个点a，找出的离这个点最远的点bi,bi一定符合，然后只需要再从这些点里拿出一个端点bk做同样操作就可以找出剩下满足的点ci。因为其他没找出来的点ci肯定是离第一次选的点a近一些，也就是离第一次找出的点bi远。所以只需要再用一个点bk就课可以找出来。
用了邻接表和深搜。
代码：
#include <iostream> #include <cstring> #include <algorithm> #define Max 10005 using namespace std; int fir[Max*2],nex[Max*2],u[Max*2],v[Max*2];//双向邻接表 int vis[Max],vi[Max];//vis标记是否访问，vi标记是否已经被认定为满足条件的点 不重复记录 int maxd[Max],no,maxv,last;//last最初为1 maxd为空，方便更新整个数组 no为下标，第一次找完后 last更新为no 方便之后的查找对no进行更新 ，不影响之前找到的点 int n,cunion;//cunion记录连通分量 void dfs(int k,int height) { if(height > maxv) { maxv = height; no = last; if(!vi[k])vi[k] = 1,maxd[no ++] = k; } else if(height == maxv && !vi[k])vi[k] = 1,maxd[no ++] = k; int p = fir[k]; while(p != -1) { if(!vis[v[p]]) { vis[v[p]] = 1; dfs(v[p],height + 1); } p = nex[p]; } } int main() { cin>>n; memset(fir,-1,sizeof(nex)); for(int i = 0;i < n - 1;i ++) { cin>>u[i]>>v[i]; u[i+n-1] = v[i]; v[i+n-1] = u[i]; nex[i] = fir[u[i]]; fir[u[i]] = i; nex[i+n-1] = fir[u[i+n-1]]; fir[u[i+n-1]] = i+n-1; } for(int i = 1;i <= n;i ++) { if(!vis[i]) { cunion ++; vis[i] = 1; dfs(i,1); } } if(cunion != 1)cout<<"Error: "<<cunion<<" components"<<endl; else { last = no; memset(vis,0,sizeof(vis)); memset(vi,0,sizeof(vi)); for(int i = 0;i < last;i ++) vi[maxd[i]] = 1; dfs(maxd[0],1); sort(maxd,maxd+no); for(int i = 0;i < no;i ++) cout<<maxd[i]<<endl; } }
标签：eof input 存在 ace void find ios scribe error
原文地址：http://www.cnblogs.com/8023spz/p/8012678.html
