PAT (Advanced Level)——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

题意：给你一个家族树,让你求每一层上叶子结点的个数。

思路：两次bfs，第一次bfs把所有点标记为在第几层上，第二个bfs判断他是否有子节点。利用邻接表，if（he[top]==0） 说明没有子节点，sum[ num[top] ]++。 sum[i]代表第i层有多少个叶子结点，num[top]代表top在第几层。

邻接表：

ll add(ll x,ll y，ll z)
{
	 ver[id]=y;//x点的终边
	 len[id]=z;// x->y的权值
	 ne[id]=he[x];//下一个节点
	 he[x]=id++;//头结点
}

满分代码：

#include<iostream>
#include<queue>
using namespace std;
typedef long long ll;
const int maxn=1e3+10;
ll num[maxn],sum[maxn],he[maxn],ne[maxn],ver[maxn];
ll id,top;
queue<ll> q1,q2;
void init()
{
	for(int i=0;i<maxn;i++)
	{
		num[i]=0;
		sum[i]=0;
		he[i]=0;
		ne[i]=0;
		ver[i]=0;
	}
}
ll bfs1(ll u)
{
	q1.push(u);
	num[u]=1;
	while(!q1.empty())
	{
		top=q1.front();
		q1.pop();
		for(int i=he[top];i;i=ne[i])
		{
			ll e=ver[i];
			num[e]=num[top]+1;
			q1.push(e);
		}
	}
}
ll bfs2(ll u)
{
	q2.push(u);
	while(!q2.empty())
	{
		top=q2.front();
		q2.pop();
		if(he[top]==0)
			sum[num[top]]++;
		else
		{
			for(int i=he[top];i;i=ne[i])
			{
				ll oo=ver[i];
				q2.push(oo);
			 } 
		}
	}
}
ll add(ll x,ll y)
{
	ver[id]=y;
	ne[id]=he[x];
	he[x]=id++;
}
int main()
{
	ll n,m;
	while(cin>>n)
	{
		init();
		cin>>m;
		id=1;
		for(int i=1;i<=m;i++)
		{
			ll root,k,a;
			cin>>root>>k;
			for(int i=1;i<=k;i++)
			{
				cin>>a;
				add(root,a);
			}
		}
		bfs1(1);
		ll m1=0;
		for(int i=1;i<=n;i++)
		{
			m1=max(m1,num[i]);
		}
		bfs2(1);
		for(int i=1;i<=m1;i++)
		{
			cout<<sum[i];
			if(i<m1)
				cout<<" ";
		}
		cout<<endl;
	}
	
	return 0;
 }