PAT-A-1094. The Largest Generation (25)
2017-05-22 23:40
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1094. The Largest Generation (25)
时间限制200 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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
#include<iostream> #include<cstdio> #include<vector> using namespace std; int gen=1, level=1; int n, m; const int genn = 110; struct node { vector<int> child; }Node[genn]; void BFS(int root) { int q[genn]; int rear = -1, front = -1; int last=0; q[++rear] = root; int temp; int l = 0; while (front < rear) { temp = q[++front]; for (int i = 0; i < Node[temp].child.size(); i++) { q[++rear] = Node[temp].child[i]; } if (last == front) { l++; last = rear; if ((rear - front)>gen) { gen = rear - front; level = l + 1; } } } } int main() { cin >> n >> m; int p, c, num; for (int i = 0; i < m; i++) { cin >> p >> num; for (int j = 0; j < num; j++) { cin >> c; Node[p].child.push_back(c); } } BFS(1); cout << gen << " " << level << endl; system("pause"); return 0; }
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