1094. The Largest Generation (25)

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题目链接：http://www.patest.cn/contests/pat-a-practise/1094

题目：

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 1321 1 2301 4 03 02 04 0503 3 06 07 0806 2 12 1313 1 2108 2 15 1602 2 09 1011 2 19 2017 1 2205 1 1107 1 1409 1 1710 1 18

Sample Output:

9 4

分析：

vector<int>...[100]来保存父节点和子节点即可，然后通过一次循环来找出最长的同辈数目

AC代码：

#include<cstdio>#include<iostream>#include<vector>#include<algorithm>#include<queue>#include<stack>#include<string>#include<string.h>using namespace std;int main(){ freopen("F://Temp/input.txt", "r", stdin); int n, m; cin >> n >> m; vector<int> vec_arr[100]; //vector<int> *vec_arr = new vector<int>[n+1];//*Point!!这个+1不能忘啊！ while (m--){  int parent,count;  cin >> parent >> count;  while (count--){   int child;   cin >> child;   vec_arr[parent].push_back(child);  } } vector<int>V1; vector<int>V2; const int ROOT = 1; V1.push_back(ROOT); int max = 1, max_level = 1, cur_level = 1; while (true){  ++cur_level;  if (V1.size() == 0)break;  for (int i = 0; i < V1.size(); ++i){   for (int j = 0; j < vec_arr[V1[i]].size(); ++j){    V2.push_back(vec_arr[V1[i]][j]);   }  }  if (V2.size() > max){   max = V2.size();   max_level = cur_level;  }  V1 = V2;  V2.clear(); } cout << max << " " << max_level << endl; //delete []vec_arr; return 0;}

截图：

——Apie陈小旭

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