A_1004

1004 Counting Leaves (30)
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

分析：最开始我是想用图来保存数据，然后从顶点1开始进行图的bfs，bfs的同时计算每行叶子节点的个数。实现起来比较复杂。看了柳婼大神的解题，才想起来使用vector。两种都是使用图来保存数据，但vector是可变长度的。而且API较多，使用起来很方便。

#include 
#define INF 999999999

using namespace std;
vector matrix[101];
int record[101] = {0};
int deepest = -1;
void dfs(int index, int depth)
{
    if(matrix[index].size() == 0){
        record[depth]++;
        deepest = deepest>depth?deepest:depth;
        return;
    }
    for(int i = 0; i < matrix[index].size(); i++)
        dfs(matrix[index][i], depth+1);
}
int main()
{
    int n, m;       //树的节点，树的非叶节点 
    cin >> n >> m;
    int id, nn, i, j, temp;
    for(i = 0; i < m; i++){
        cin >> id >> nn;
        for(j = 0; j < nn; j++){
            cin >> temp;
            matrix[id].push_back(temp);
        }
    }
    dfs(1, 0);
    cout << record[0];
    for(i = 1; i <= deepest; i++)
        cout << " " << record[i];
    return 0;
}