PAT (Advanced Level) 1004. Counting Leaves (30) 层序遍历，两个queue辅助

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

Input

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.

Output

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

2 101 1 02

Sample Output

0 1

使用一个二维数组，第i个元素存储i结点的所有子节点序号。层序遍历得到结果。
/*2015.7.19cyq*/#include <iostream>#include <vector>#include <queue>using namespace std;int main(){int N,M;cin>>N>>M;vector<vector<int> > tree(N+1);//tree[i]记录i结点的子结点集合int id,k,x;while(M--){cin>>id>>k;while(k--){cin>>x;tree[id].push_back(x);}}vector<int> result;queue<int> cur,next;cur.push(1);while(!cur.empty()){int level=0;while(!cur.empty()){int root=cur.front();cur.pop();if(tree[root].empty())//无子结点，说明root为叶子结点level++;for(auto it=tree[root].begin();it!=tree[root].end();it++){next.push(*it);}}result.push_back(level);swap(cur,next);}cout<<result[0];for(auto it=result.begin()+1;it!=result.end();++it)cout<<" "<<*it;return 0;}