PTA甲级1094 The Largest Generation (25分)

PTA甲级1094 The Largest Generation (25分)

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

【程序思路】

定义一个结构体，保存每一行输入的数据。结构体中的height记录该节点所在的高度。
然后利用队列层序遍历树，定义数组c，索引为高度，值为该高度的节点个数。最后遍历数组c找到最大值的下标，再输出。

【程序实现】

#include<bits/stdc++.h>
using namespace std;
struct tree{
    int *child = NULL;
    int height, k;
}Tree[105] , t;
int main(){
	int n, m, x, k, c[105] = {0 , 1}, index = 0;
    queue<struct tree> q;
    cin>>n>>m;
    for(int i = 0; i < m; i++) {
        cin>>x>>k;
        Tree[x].child = new int[k];
        Tree[x].k = k;
        for(int j = 0; j < k; j++)
            cin>>Tree[x].child[j];
    }
    Tree[1].height = 1;
    q.push(Tree[1]);
    while(!q.empty()) {
        t = q.front();
        q.pop();
        if (t.child != NULL) {
            for (int i = 0; i < t.k; i++) {
                Tree[t.child[i]].height = t.height + 1;
                c[t.height + 1] ++;
                q.push(Tree[t.child[i]]);
            }
        }
    }
    for (int i = 1; i < 105; i++)
        if (c[i] > c[index])
            index = i;
    cout<<c[index]<<' '<<index;
	return 0;
}