Even Tree 小议

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/chr1991/article/details/80945436

原题链接：https://www.hackerrank.com/challenges/even-tree/problem

思路：把树还原出来，对每个结点，计算以该结点为根的子树结点数。子树结点数为偶数的结点数量就是所求森林中树木的数量。结点数量减去一就是所求结果。

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

struct node{
    int index, num_subtree;
    vector<node*> children;
    node(int i=0,int n=0):
        index(i),num_subtree(n), children(vector<node*>()){}
};

node* construct_tree(int parent, int index, const vector<vector<int>>& adjLsit);
void count_even(node* root, int& cnt);

int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */
    int N, M; cin >> N >> M;
    vector<vector<int>> adjList(N + 1);
    for(int i = 0; i < M; i++){
        int n1, n2; cin >> n1 >> n2;
        adjList[n1].push_back(n2);
        adjList[n2].push_back(n1);
    }
    node* root = construct_tree(0, 1, adjList);
    int cnt = 0;
    count_even(root, cnt);
    cout << cnt - 1 << endl;
    return 0;
}

node* construct_tree(int parent, int index, const vector<vector<int>>& adjList){
    node* tmp = new node(index);
    tmp->num_subtree = 1;
    for(int i = 0; i < adjList[index].size(); i++){
        if(adjList[index][i] != parent){
            tmp->children.push_back(construct_tree(index, adjList[index][i], adjList));
            tmp->num_subtree += tmp->children.back()->num_subtree;
        }
    }
    return tmp;
}

void count_even(node* root, int& cnt){
    for(auto vit: root->children){
        count_even(vit, cnt);
    }
    if(root->num_subtree%2==0){
        cnt++;
    }
}