1094 The Largest Generation (25分)

最新推荐文章于 2024-06-02 16:33:44 发布

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最新推荐文章于 2024-06-02 16:33:44 发布

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本文介绍了一种通过深度优先搜索（DFS）算法遍历家族树结构的方法，旨在找出具有最多成员的一代及其具体代数。通过对输入数据进行解析，构建家族成员间的连接，算法能够有效地确定各代的人口数量，并最终输出人口最多的代数及其人口数。

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A family hierarchy is usually presented by a pedigree tree where all the nodes on the same level belong to the same generation. Your task is to find the generation with the largest population.

Input Specification:

Each input file contains one test case. Each case starts with two positive integers N (<100) which is the total number of family members in the tree (and hence assume that all the members are numbered from 01 to N), and M (<N) which is the number of family members who have children. Then M lines follow, each contains the information of a family member in the following format:

ID K ID[1] ID[2] ... ID[K]

where ID is a two-digit number representing a family member, K (>0) is the number of his/her children, followed by a sequence of two-digit ID's of his/her children. For the sake of simplicity, let us fix the root ID to be 01. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the largest population number and the level of the corresponding generation. It is assumed that such a generation is unique, and the root level is defined to be 1.

Sample Input:

23 13
21 1 23
01 4 03 02 04 05
03 3 06 07 08
06 2 12 13
13 1 21
08 2 15 16
02 2 09 10
11 2 19 20
17 1 22
05 1 11
07 1 14
09 1 17
10 1 18

Sample Output:

9 4

题意：建立一棵祖先树，每一层代表一代人，求最多那代人的人数和是哪一代人，根为第一代。

思路：用DFS遍历的同时使用height来保存第几代，并使用数组记录每代人数（下标为第几代），最后遍历数组，求出最大值并输出值和下标。

#include<iostream>

#include<vector>

using namespace std;
vector<int> vec[100];
bool visit[100] = {false};
int cnt_gen[100];
void dfs(int start, int height){

    cnt_gen[height]++;

    for(int i = 0; i < vec[start].size(); i++){

        if(!visit[vec[start][i]]){
            
            visit[vec[start][i]] = true;
            dfs(vec[start][i], height + 1);
        }
    }

}
int main(){

    int n, m;
    cin >> n >> m;

    for(int i = 0; i < m; i++){
        int id, k;
        scanf("%d %d", &id, &k);
        for(int j = 0; j < k; j++){
            int temp;
            scanf("%d", &temp);
            vec[id].push_back(temp);
        }

    }

    dfs(1, 1);
    int max_cnt = -1;
    int max_level = -1;
    for(int i = 1; i < 100; i++){
        if(cnt_gen[i] == 0){
            break;
        }
        if(max_cnt < cnt_gen[i]){
            max_cnt = cnt_gen[i];
            max_level = i;
        }
    }
    printf("%d %d\n", max_cnt, max_level);
    return 0;
}