PAT A1004 Counting Leaves （30 分）—

PAT A1004 Counting Leaves （30 分）——树，DFS，BFS

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 $0, the number of nodes in a tree, and$

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

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 01level, 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

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
#include <iostream>
using namespace std;
int tree[101][101] = { -1 };//只会赋给第一个元素，别的默认为0
//fill(tree[0], tree[0] + 101*101, -1);
int leaf[101] = { 0 };
int step = 0;
void dfs(int i, int depth){
    if (tree[i][0] == -1){
        leaf[depth]++;
        return;
    }
    int j = 0;
    while (tree[i][j] != -1){
        dfs(tree[i][j], depth + 1);
        j++;
    }
    step = max(step, depth + 1);
}
int main(){
    int n, m;
    fill(tree[0], tree[0] + 101*101, -1);
    cin >> n >> m;
    for (int i = 0; i < m; i++){
        int root, num;
        cin >> root >> num;
        for (int j = 0; j < num; j++){
            int leaf;
            cin >> leaf;
            tree[root][j] = leaf;
        }
    }
    int root = 1, depth = 0;
    dfs(root, depth);
    for (int i = 0; i <= step; i++){
        if(i!=step)cout << leaf[i] << " ";
        else cout << leaf[i];
    }
    system("pause");
}

注意点：一开始题目都没看懂，给的例子太不具代表性了，读了好多遍终于知道是要求每层的叶节点个数并输出。就是考了一个深度优先搜索（DFS），居然还不会做，DFS要用递归一层层遍历，遍历完要记录最深到几层了，方便后面输出。这里建树用了二维数组，其实有点蠢，用vector数组会方便一点。另外，二维数组的初始化只能用fill或memset，直接像一维数组那样给一个值只会赋给数组第一个元素。

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