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[100];
int main(){
    int n,m;
    scanf("%d %d",&n,&m);
    int i,num,count,in;
    for(i=0;i<m;i++){
        scanf("%d %d",&num,&count);
        while(count--){
            scanf("%d",&in);
            v[num].push_back(in);
        }
    }
    queue<int> q;
    q.push(1);
    int last=1,tail,floor=1,cur,maxcount=1,maxfloor=1,curcount=0;
    while(!q.empty()){
        cur=q.front();
        q.pop();
        for(i=0;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=0;
        }
    }
    printf("%d %d
",maxcount,maxfloor);
    return 0;
}