1004 Counting Leaves （PAT甲级）

#include 
#include 
const int maxN = 100;

int N, M, id, K, level, p;
int nbr[maxN];
std::vector vec[maxN];

void dfs(int k){
    if(vec[k].empty()){
        nbr[level]++;
        return;
    }
    level++;
    for(int i = 0; i < vec[k].size(); ++i){
        dfs(vec[k][i]);
    }
    level--;
}

int main(){
    scanf("%d %d", &N, &M);
    for(int i = 0; i < M; ++i){
        scanf("%d %d", &id, &K);
        vec[id].resize(K);
        for(int j = 0; j < K; ++j){
            scanf("%d", &vec[id][j]);
        }
    }
    level = 0;
    dfs(1);
    for(int i = 0; i < maxN; ++i){
        if(nbr[i] != 0){
            p = i;
        }
    }
    for(int i = 0; i <= p; ++i){
        printf("%d%s", nbr[i], i == p ? "\n" : " ");
    }
    return 0;
}

题目如下：

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

Input Specification:

Each input file contains one test case. Each case starts with a line containing 0

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.

The input ends with N being 0. That case must NOT be processed.

Output Specification:

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