PAT_1094

1094 The Largest Generation

题目描述

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:

1
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:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
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:

1
9 4

思路

给出树中每个节点的子节点,并且根节点为01,求解树中包含节点最多的一层以及节点数量

在跟节点开始,BFS向深处遍历,即可得到每一层包含的节点,判断出包含最多的一层即可

程序

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#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <set>
#include <queue>
#include <map>
#define INF 0x3f3f3f3f

using namespace std;

const int maxn = 110;
const int mod = 1000000007;
typedef long long ll;
vector<int> ch[maxn];
int n,m,num[maxn],maxdeep;
struct Node
{
int pos,deep;
};

void BFS()
{
Node node{1,1},tmp;
queue<Node> que;
que.push(node);
while(!que.empty())
{
node = que.front();que.pop();
num[node.deep] += 1;
maxdeep = max(maxdeep,node.deep);
for(int i = 0;i < ch[node.pos].size();i ++)
{
tmp.pos = ch[node.pos][i];
tmp.deep = node.deep + 1;
que.push(tmp);
}
}
}
int main()
{
int id,k,x;
scanf("%d%d",&n,&m);
for(int i =0 ;i < m;i ++)
{
scanf("%d%d",&id,&k);
for(int j =0 ;j < k;j ++)
{
scanf("%d",&x);
ch[id].push_back(x);
}
}
BFS();
int a = 0,b;
for(int i =0 ;i <= maxdeep;i ++)
if(num[i] > a)
a = num[i],b = i;
printf("%d %d\n",a,b);

return 0;
}
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