1094 The Largest Generation
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
题目大意:让你统计一颗树每层节点的最大数目,以及相应层次。
大致思路:用BFS统计层序遍历每一层节点,同时定义一个数组level用来记录每一层结点的高度,其中level(孩子节点) = level(父节点) + 1。定义一个数组cnt用来统计每一层结点的个数。
代码:
#include <bits/stdc++.h>
using namespace std;
const int N = 110;
vector<int> root[N];
int n, m;
int ans1, ans2;
int level[N], cnt[N]; //统计每一层树的高度和每一层的节点数
void BFS(int x) {
queue<int> q;
q.push(x);
level[1] = 1;
ans1 = 1, ans2 = 1;
while (!q.empty()) {
int t = q.front();
q.pop();
cnt[level[t]]++;
// cout << cnt[level[t]] << endl;
for (int i = 0; i < root[t].size(); i++) {
level[root[t][i]] =
level[t] + 1; //孩子结点的层数等于父结点层数 + 1
q.push(root[t][i]);
}
}
for (int i = 1; i <= n; i++) {
// cout << cnt[level[i]] << endl;
if (ans1 < cnt[level[i]]) {
ans1 = cnt[level[i]];
ans2 = level[i];
}
}
cout << ans1 << " " << ans2 << endl;
}
int main() {
scanf("%d %d", &n, &m);
memset(level, 0, sizeof(level));
memset(cnt, 0, sizeof(cnt));
for (int i = 0; i < m; i++) {
int id, k;
scanf("%d %d", &id, &k);
for (int j = 0; j < k; j++) {
int x;
scanf("%d", &x);
root[id].push_back(x);
}
}
BFS(1);
return 0;
}