1094. The Largest Generation (25)

1094. The Largest Generation (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

发现这道其貌不扬的题目是上次PAT甲级的原题，然后就大发慈悲重新写了一下，上次做的时候我还不会用stl，这次写起来稍微简练一点了。给你一张图，标出了一个家族内部孩子和父亲的关系，然后希望你找出人数最多的一代，很容易想到广度优先搜索，然后统计每一层的人数。找出最大值。找出最大值最便捷的方式是使用priority_que。
更简单的方法是把这个看成一张图，直接使用Floyd算法找出各个节点到根节点的个数。不过我没有试过，不知道会不会超时。之后扫一遍统计个数就好了。

# include <cstdio>
# include <queue>
# include <vector>
using namespace std;

const int size = 100;

vector<int> v[size];
typedef pair<int,int> sum_log;
priority_queue<sum_log> pri_que; /*in order to find the max one*/

void bfs(int root)
{
queue<int> que;
int i = 1,j;
que.push(root);
while (!que.empty())
{
queue<int> tmp;
pri_que.push(sum_log(que.size(),i++));
while (!que.empty())
{
int loca = que.front();que.pop();
for (j=0;j<v[loca].size();j++)
tmp.push(v[loca][j]);
}
que = tmp;
}
}
int main()
{
int i,j,k,tmp;
int n,m;
scanf("%d%d",&n,&m);
while (m--)
{
scanf("%d%d",&i,&j);
while (j--)
{
scanf("%d",&k);
v[i].push_back(k);
}
}
bfs(1);
sum_log ans = pri_que.top();
printf("%d %d\n",ans.first,ans.second);
return 0;
}