PAT (Advanced Level) Practice 1094 The Largest Generation (25 分)樹的層次遍歷
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
輸出樹中最多節點的個數和所在的層數。
程式碼如下:
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
const int maxn=105;
int n,m;
vector<int>v[maxn];
int par[maxn];
int height,num;
void traverse (int x)
{
queue<int>q;
height=1; num=1;
q.push(x);
int k=0;
while (!q.empty())
{
k++;
int Size=q.size();
if(num<Size)
{
num=Size;
height=k;
}
while (Size--)
{
int t=q.front();
q.pop();
for (int i=0;i<v[t].size();i++)
{
q.push(v[t][i]);
}
}
}
}
int main()
{
scanf("%d%d",&n,&m);
memset (par,-1,sizeof(par));
for (int i=0;i<m;i++)
{
int x,num;
scanf("%d%d",&x,&num);
for (int j=0;j<num;j++)
{
int y;
scanf("%d",&y);
v[x].push_back(y);
par[y]=x;
}
}
int root;
for (int i=1;i<=n;i++)
if(par[i]==-1)
{
root=i;
break;
}
traverse(root);
printf("%d %d\n",num,height);
return 0;
}