PTA甲级-1004 Counting Leaves （30 分）

1004 Counting Leaves （30 分）

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 0<N<100, the number of nodes in a tree, and M (<N), the number of non-leaf nodes. Then M lines follow, each in the format:

ID K ID[1] ID[2] … ID[K]

where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID’s of its children. For the sake of simplicity, let us fix the root ID to be 01.

The input ends with N being 0. That case must NOT be processed.

Output Specification:

For each test case, you are supposed to count those family members who have no child for every seniority level starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.

The sample case represents a tree with only 2 nodes, where 01 is the root and 02 is its only child. Hence on the root 01 level, there is 0 leaf node; and on the next level, there is 1 leaf node. Then we should output 0 1 in a line.

Sample Input:

2 1
01 1 02

Sample Output:

0 1

30分究极水水水水水题= = 太水了

题目意思：先给你n m两个数
n代表一共有多少节点，m表示下面多少行
然后下面每一行是这样的：

爹的ID k个数 儿子1 儿子2 儿子3 … 儿子k

然后让你求 每一层有多少叶子节点（也就是没儿子的爹 ）
思路很简单：
标记好父亲（pre数组） 标记好儿子（son数组） 然后用queue 层序遍历每一层
把每一层的叶子节点放进vector中
最后顺序输出好了（真正的送分题 ）
这题主要是要记录每一层要pop多少次，毕竟是多叉树，哈哈

#include <bits/stdc++.h>
const int maxn=5e5+10;
const int mod=1e9+7;
#define INF 2147483647
#define PI atan(1)*4
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define ss(a) scanf("%s",&a)
#define sc(a) scanf("%d",&a)
#define scc(a,b) scanf("d",&a,&b)
#define sccc(a,b,c) scanf("d%d",&a,&b,&c)
using namespace std;
int pre[105];
int son[105];
queue <int> cen;
vector <int> ans;
int n,m,p,s,a;
int main()
{
    while(cin>>n>>m)
    {
        while(!cen.empty())
            cen.pop();
        CL(son,0);
        for(int i=0;i<=100;i++)
            pre[i]=i;
        //初始化
        while(m--)
        {
            cin>>s>>p;
            for(int i=0;i<p;i++)
            {
                cin>>a;
                pre[a]=s;//a的父亲是s
                son[s]++;//s这个父亲有儿子
            }
        }
        int len=1;//这个表示要pop多少次
        cen.push(1);
        while(!cen.empty())
        {
            int flag=0;//这个表示pop到哪里了
            int tt=0;
            while(len>flag)
            {
                int now=cen.front();//记录当前队列头
                flag++;
                for(int i=1;i<=n;i++)
                    if(i!=now && pre[i]==now)
                        cen.push(i);
                if(!son[now])tt++;
                cen.pop();
            }
            len=cen.size();//记录这层的元素个数（要pop多少次）
            ans.push_back(tt);//这层叶子节点一共有几个
        }
        for(int i=0;i<ans.size();i++)
        {
            if(i!=0)printf(" ");
            printf("%d",ans[i]);
        }
        printf("\n");
    }
    return 0;
}