PAT甲级1004Counting Leaves

1004. Counting Leaves (30)
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 1
01 1 02
Sample Output
0 1

这题目挺难读懂的。。我理解了很久终于发现原来只需要判断一棵树每层有多少个叶子节点就好了。
既然知道了题目意思，那么就直接dfs就好了。这里的思路是记录下dfs每次到达节点的层数，再依据当前节点的子节点有无进行操作。只要所有节点都被遍历一次那么答案也就出来了。

#include 
#include
#include 
#include
using namespace std;
vector<int>asd[102];
int ans[102];
int ma;
void dfs(int now,int cen){
    int sum = 0;
    for(int i=0;i1);
        sum++;
    }
    if(sum==0)
    ans[cen]++;
    ma=max(cen,ma);
}

int main(){
    int m,n;
    cin>>n>>m;
    for(int i=1;i<=m;i++){
        int a,b,c;
        cin>>a>>b;
        for(int j=0;jcin>>c;
            asd[a].push_back(c);
        }
    }
    memset(ans,0,sizeof(ans));
    dfs(1,1);
    for(int i=1;iprintf("%d ",ans[i]);
    cout<return 0;
}