PAT(Advanced)1004 Counting Leaves (30 分)

1004 Counting Leaves (30 分)

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 $0<N<100$, 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.

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

题目解读：输入一串树的节点，及这些节点的孩子的编号，01为跟节点，输出每一层的叶节点个数。

解题思路：用vector<int> T[100]，就是一个二维数组储存叶节点和孩子的编号，有点像存图的边。这道题方便的就是01一定是根节点，这样就可以DFS得到每个节点的高度，把没有孩子的节点按层记数，因为中间可能有很多0，所以记录一下最深层。这次卡在了递归里的++deep，deep++每一次都会给deep+1，但其实我要的就是deep+1，慎用++的操作。

AC代码

#include <iostream>
#include <vector>
using namespace std;
int leave[100];
vector<int> T[100];
int n,m;
int maxdep = 0;
void DFS(int root,int deep){
    if(T[root].size() == 0 )leave[deep]++;
    if(maxdep<deep)
        maxdep = deep;
    for( int i = 0 ; i < T[root].size() ; i ++ ){
        DFS(T[root][i],deep+1);
    }
}

int main(int argc, const char * argv[]) {
    cin >> n >> m;
    int ID,k,c;
    for( int i = 0 ; i < m ; i ++ ){
        cin >> ID >> k;
        for( int j = 0 ; j < k; j ++ ){
            cin >> c;
            T[ID].push_back(c);
        }
    }
    DFS(1,0);
    for( int i = 0 ; i < maxdep ; i ++ ){
        cout << leave[i] << " ";
    }
    cout << leave[maxdep];
    //std::cout << "Hello, World!\n";
    return 0;
}