PAT-A | 1094 | The Largest Generation

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

解题思路：

每个结点都保存它的父结点和它的层数。

1. 处理输入。根据输入的格式，每一行输入都可以知道子节点的父节点的编号，将其记录下来，level初始化为0.

2. 计算层数。从根结点开始，层序遍历，用队列实现。首先计算根结点的子节点的层数，然后一层一层向下推进。

3. 计算结点数。上一步知道了每个结点的层数，对每个结点遍历一次，统计每一层的结点数，保存在数组gen[]中。

4. 输出结果。知道了每一层的结点数，找出结点数最大的那一层，输出该层结点数与该层号即可。

#include <iostream>
#include <queue>

using namespace std;

struct node {
int level;
int parent;
};
typedef node *tree;

int main()
{

int N, M;
int pid, K, cid; //parent id, K, child id
tree T;
int *gen; // nodes of every generation
queue<int> Q;
int f;//front of queue

cin >> N >> M;

T = new node[N + 1];
gen = new int[N + 1];

//step 1. process input
for (int i = 0; i != M; ++i) {
cin >> pid >> K;
for (int j = 0; j != K; ++j) {
cin >> cid;
T[cid].parent = pid;
T[cid].level = 0; //initialize
}
}

//step 2. calculate every node's level
T[1].level = 1;//root
Q.push(1);//root
while (!Q.empty()) {
f = Q.front(); Q.pop();
for (int i = 1; i != N + 1; ++i) {
if (T[i].parent == f) {
T[i].level = T[f].level + 1;
Q.push(i);
}
}
}

//debug
//for (int i = 1; i != N + 1; ++i) {
//	cout << i << " " << T[i].parent << " " << T[i].level << endl;
//}

//step 3. calculate the number of nodes of each level

//initialize gen[]
for (int i = 1; i != N + 1; ++i) {
gen[i] = 0;
}

int maxLevel = 1;
for (int i = 1; i != N + 1; ++i) {
++gen[T[i].level];
if (T[i].level > maxLevel) {
maxLevel = T[i].level;
}
}

//debug
//for (int i = 1; i <= maxLevel; ++i) {
//	cout << gen[i] << " ";
//}

//step 4. find the largest population number and the level of the corresponding generation
int maxNodes = 0;
int level = 0;
for (int i = 1; i <= maxLevel; ++i) {
if (gen[i] > maxNodes) {
maxNodes = gen[i];
level = i;
}
}

cout << maxNodes << " " << level << endl;

delete[]T;
delete[]gen;
}

这题与PAT-A 1004 Counting Leaves 类型相同。