PAT 1004. Counting Leaves (30)

1004. Counting Leaves (30)

时间限制

400 ms

内存限制

32000 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. 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

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提交代码

//如果对你有帮助，希望回复下，谢谢！#include<stdio.h>const int MAX=101;int parent[MAX],level[MAX],isLeaf[MAX],maxLevel=0;int findLevel(int x){if(parent[x]==-1)return 1;elsereturn findLevel(parent[x])+1; //这是个递归过程!} int main(){int n,m,tmp,id,num,i;//freopen("G:\\in.txt","r",stdin);while(scanf("%d%d",&n,&m)!=EOF){for(i=1;i<=n;i++){parent[i]=-1;   //务必初始化，不然超时。。。。。。isLeaf[i]=1;level[i]=0;}while(m--){scanf("%d%d",&tmp,&num);//输入为：父节点 以及 孩子的个数while(num--){scanf("%d",&id);parent[id]=tmp;}isLeaf[tmp]=0;}for(i=1;i<=n;i++){if(isLeaf[i]){tmp=findLevel(i);//找到叶子结点所在的层，根记为第1层。level[tmp]++;if(tmp>maxLevel)maxLevel=tmp;}}for(i=1;i<=maxLevel;i++){if(i!=1) printf(" ");printf("%d",level[i]);}printf("\n");}return 0;}

比较好的代码：
http://blog.csdn.net/jjike/article/details/8615773
http://blog.csdn.net/huntinggo/article/details/18943691
http://blog.csdn.net/cstopcoder/article/details/19018767