1021 Deepest Root (25)（25 point(s)）

problem

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

tip

answer

#include <cassert>
#include <set>
#include <iostream>
#include <vector>
#include <cstring>

#define Max 10010
#define fi first
#define se second
using namespace std;

int N;
int Root[Max], V[Max];
vector<int > E[Max];
set<int > S, Last;
set<int>::iterator it;

void Init(){
    for(int i = 1; i <= N; i++){
        Root[i] = i;
    }
}

int Find(int i){
    if(Root[i] != i) Root[i] = Find(Root[i]);
    return Root[i];
}

void Union(int a, int b){
    int ra = Find(a);
    int rb = Find(b);
    if(ra == rb) return ;
    Root[ra] = Find(Root[rb]);
}

int TreeNum() {
    int num = 0;
    for(int i = 1; i <= N; i++){
        if(Root[i] == i) num++;
    }
    return num;
}

int DFS(int t, int deep, int &maxD){
    if(deep > maxD){
        maxD = deep;
        S.clear();
        S.insert(t);
    }
    if(deep == maxD){
        S.insert(t);
    }
    V[t] = 1;
    for(int i = 0; i < E[t].size(); i ++){
        if(!V[E[t][i]]){
            DFS(E[t][i], deep+1, maxD);
        }
    }
}

void InitV(){
    memset(V, 0, sizeof(V));
}

void PrintStatus(){
    for(it = S.begin(); it != S.end(); it++){
        cout<<*it<<" ";
    }
    cout<<endl;
}
int main(){
//  freopen("test.txt", "r", stdin) ;
    ios::sync_with_stdio(false);
    //N<= 10000
    cin>>N;
    if(N == 0){
        assert(false);
        cout<<"0"<<endl;
    }
    Init();
    int a, b;
    for(int i = 0; i < N-1; i++) {
        cin>>a>>b;
        E[a].push_back(b);
        E[b].push_back(a);
        Union(a, b);
    }
    if(TreeNum() > 1) {
        cout<<"Error: "<<TreeNum()<<" components";
    }else{
        int maxD = 0, deepN;
        DFS(1, 0, maxD);
//      PrintStatus(); 
        for(it = S.begin(); it != S.end(); it++){
            Last.insert(*it);
            deepN = *it;
        }
        maxD = 0;
//      cout<<deepN<<endl;
        InitV();
        DFS(deepN, 0, maxD);
//      PrintStatus();
        for(it = S.begin(); it != S.end(); it++){
            Last.insert(*it);
        }
        for(it = Last.begin(); it != Last.end(); it++){
            cout<<*it<<endl;
        }
    }
    return 0; 
}

exprience

• acyclic 無環的