PAT甲级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

求树的最长路（直径）问题：
1.用邻接表储存题目所给的图；
2.用并查集判断连通分量的个数，如果大于1，就不是树，输出错误信息和连通分量的个数。
3.如果是树，就求该树的直径：从树上的任意结点v出发，用DFS/BFS找到距v最远的点w（可能不止一个），w 即是直径上的一个端点；再以 w 为起点再次DFS/BFS，可以找到直径上的其他端点。

> 证明：
设树上有一条直径ST。需要证明的是：从树上的任意结点v出发，找到的w一定是直径的端点。
分情况讨论：
1）如果v在ST路径上，如果找到的w不是S或T，即
=>distance(v,w)>distance(v,S) 和 distance(v,w)>distance(v,T)
=> distance(v,w) + distance(v,T) > distance(v,S) +distance(v,T)
即路径wT的距离大于ST，与ST是直径的前提矛盾。
2）如果v不在ST路径上，
a.如果vw和ST有交点X，v首先走到X，从X出发，找到的w一定是S或者T，这一点在1）中已经证明过了；
b.如果vw和ST没有交点，即
distance(v,w) > distance(v,X) + distance(X,S) 和 distance(v,w) > distance(v,X) + distance(X,T),
其中X是vS和vT路径的交点。
=>distance(v,w) + distance(v,X)+distance(X,T) > distance(X,S) + distance(X,T)
情况1

情况2-a

情况2-b

参考资料
http://stackoverflow.com/questions/25649166/linear-algorithm-of-finding-tree-diameter
http://www.xuebuyuan.com/225744.html

#include <cstdio>using namespace std;#include <set>#include <cstring>#include <vector> const int MAX=10000+10;vector<int> Node[MAX];//邻接表int vis[MAX]={0};int maxHeight=0;int currentHeight=0;vector<int> tmp;set<int> ans;int N;//结点数int cnt; //连通分量数int P[MAX];int Find(int x){    if(P[x]<0) return x;    return x=Find(P[x]);}void Union(int x,int y){    int px=Find(x),py=Find(y);    if(px!=py) {        if(-P[px]>-P[py]) {            P[px]+=P[py];            P[py]=px;        }        else{            P[py]+=P[x];            P[px]=py;        }        cnt--;    }}void DFS(int index){    vis[index]=1;    if(currentHeight>maxHeight) {        maxHeight=currentHeight;        tmp.clear();        tmp.push_back(index);    }    else if(currentHeight==maxHeight){        tmp.push_back(index);    }    for(int i=0;i<Node[index].size();i++){        if(vis[Node[index][i]]==0) {            currentHeight++;            DFS(Node[index][i]);            currentHeight--;        }    }}int main(){    scanf("%d",&N);    cnt=N;    for(int i=1;i<=N;i++) P[i]=-1;    int u,v;    for(int i=0;i<N-1;i++){        scanf("%d %d",&u,&v);        Node[u].push_back(v);        Node[v].push_back(u);        Union(u,v);    }    if(cnt>1) {        printf("Error: %d components\n",cnt);       }    else{        DFS(1);        memset(vis,0,sizeof(vis));        currentHeight=0;        for(int i=0;i<tmp.size();i++) ans.insert(tmp[i]);        DFS(tmp[0]);            for(int i=0;i<tmp.size();i++) ans.insert(tmp[i]);        set<int>::iterator it;        for(it=ans.begin();it!=ans.end();it++) {            printf("%d\n",*it);        }    }    return 0;}