1004. Counting Leaves (30)

题目链接：https://www.patest.cn/contests/pat-a-practise/1004
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 1
01 1 02
Sample Output
0 1
题意解析：对一个普通的树，求每层的叶子节点个数

#include<cstdio>#include<vector>#include<algorithm>using namespace std;const int MAXN=110;vector<int> G[MAXN];//存放树 int leaf[MAXN]={0};//存放每层的叶子结点数 int max_h=1;//树的深度 void DFS(int index,int h){//index为当前遍历到的节点编号，h为当前深度     max_h=max(h,max_h);    if(G[index].size()==0){//如果该节点为叶子节点         leaf[h]++;        return;    }    for(int i=0;i<G[index].size();i++){//枚举所有子节点         DFS(G[index][i],h+1);//对每层进行深度优先递归    }}int main(){    int n,m,parent,child,k;//n是节点总数，m是非叶子节点数     scanf("%d %d",&n,&m);    for(int i=0;i<m;i++){        scanf("%d%d",&parent,&k);        for(int j=0;j<k;j++){            scanf("%d",&child);            G[parent].push_back(child);                 }    }    DFS(1,1);    printf("%d",leaf[1]);    for(int i=2;i<=max_h;i++){        printf(" %d",leaf[i]);    }    return 0;}