pat1094. 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

---

提交代码

姥姥教的方法，顺便用一下。

last：更新前，本层最后一个元素
tail：下一层当前最后一个元素
floor：指向当前的层数

 #include<cstdio>
 #include<algorithm>
 #include<iostream>
 #include<cstring>
 #include<queue>
 #include<vector>
 #include<cmath>
 using namespace std;
 vector<int> v[];
 int main(){
     int n,m;
     scanf("%d %d",&n,&m);
     int i,num,count,in;
     for(i=;i<m;i++){
         scanf("%d %d",&num,&count);
         while(count--){
             scanf("%d",&in);
             v[num].push_back(in);
         }
     }
     queue<int> q;
     q.push();
     int last=,tail,floor=,cur,maxcount=,maxfloor=,curcount=;
     while(!q.empty()){
         cur=q.front();
         q.pop();
         for(i=;i<v[cur].size();i++){
             q.push(v[cur][i]);
             tail=v[cur][i];
             curcount++;
         }
         if(last==cur){
             last=tail;
             floor++;//point to next floor
             if(curcount>maxcount){
                 maxcount=curcount;
                 maxfloor=floor;
             }
             curcount=;
         }
     }
     printf("%d %d\n",maxcount,maxfloor);
     return ;
 }