• 1094. The Largest Generation (25)


    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
     1 #include<string>
     2 #include<iostream>
     3 #include<map>
     4 #include<stdio.h>
     5 #include<vector>
     6 using namespace std;
     7 
     8 struct node
     9 {
    10     vector<int> child;
    11 };
    12 
    13 node Tree[110];
    14 int Level[110];
    15 int MAX = -1;
    16 void DFS(int n,int level)
    17 {
    18     if(MAX < level)
    19         MAX = level;
    20     ++Level[level];
    21     for(int i = 0 ;i < Tree[n].child.size() ;++i)
    22         DFS(Tree[n].child[i],level+1);
    23 }
    24 int main()
    25 {
    26     int n,m,tid,num,id;
    27     scanf("%d%d",&n,&m);
    28     for(int i = 0 ;i < m ;++i)
    29     {
    30         scanf("%d%d",&id,&num);
    31         for(int k = 0;k < num;++k)
    32         {
    33             scanf("%d",&tid);
    34             Tree[id].child.push_back(tid);
    35         }
    36     }
    37     DFS(1,1);
    38     int maxL = -1;
    39     int index = -1;
    40     for(int i = 1; i <= MAX;++i)
    41     {
    42         if(Level[i] > maxL)
    43         {
    44             maxL = Level[i];
    45             index = i;
    46         }
    47     }
    48     printf("%d %d
    ",maxL,index);
    49     return 0;
    50 }
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  • 原文地址:https://www.cnblogs.com/xiaoyesoso/p/5210943.html
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