1094. The Largest Generation (25)

迈不过友情╰ 2022-05-30 03:41 128阅读 0赞

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

--------------------

题目大意：

代码：

#include<stdio.h>
    #include<vector>
    #include<queue>
    using namespace std;
    struct node
    {
        int level;
        vector<int> v;
    }person[101];
    int cnt[101];
    int main()
    {
        int i,j,n,m,k,t,id,tmp;
        scanf("%d %d",&n,&m);
        for(i=0;i<m;i++)
        {
            scanf("%d %d",&k,&t);
            for(j=0;j<t;j++)
            {
                scanf("%d",&tmp);
                person[k].v.push_back(tmp);
            }
        }
        queue<int> q;
        q.push(1);
        person[1].level=1;
        while(!q.empty())
        {
            int num=q.front();
            q.pop();
            cnt[person[num].level]++;
            for(i=0;i<person[num].v.size();i++)
            {
                q.push(person[num].v[i]);
                person[person[num].v[i]].level=person[num].level+1;
            }
        }
        int Max=0,index;
        for(i=1;i<100;i++)
        {
            if(cnt[i]>Max)
            {
                Max=cnt[i];
                index=i;
            }
        }
        printf("%d %d\n",Max,index);
        return 0;
    }