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 Input2 1 01 1 02Sample Output
0 1
#include<cstdio>
#include<vector>
#include<deque>
#define MAX 105
using namespace std;
typedef struct Node
{
int child[MAX];//孩子节点的下标
int num;//孩子的个数
}Node;
vector<Node> v(MAX);//用下标表示节点编号
deque<int> q(MAX*MAX);
int main(int argc,char *argv[])
{
int N,M;
int i,j,k;
scanf("%d%d",&N,&M);
for(i=1;i<=N;i++)
v[i].num=0;
for(i=1;i<=M;i++)
{
int ID;
scanf("%d%d",&ID,&k);
for(j=1;j<=k;j++)
scanf("%d",&(v[ID].child[j]));
v[ID].num=k;
}
q[0]=1;
k=1;
for(int l=0,r=1,t=0;l<k;r=k,t++)
{//t表示树的层数，l表示第t层最左边节点的编号，r表示第t层最右边节点的编号
int count=0;
for(;l<r;l++)//每一层节点操作
{
int index=q[l];
if(!v[index].num)
count++;
for(i=1;i<=v[index].num;i++)
q[k++]=v[index].child[i];//孩子节点入队列
}
if(t)
printf(" %d",count);
else
printf("%d",count);
}
printf("\n");
return 0;
}
Pat(Advanced Level)Practice--1004(Counting Leaves)
原文：http://blog.csdn.net/u012736084/article/details/19018767