PTA 1004 Counting Leaves (30 分)（详细翻译+注释）

怼烎@ 2022-09-04 03:45 324阅读 0赞

# 背景： #

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

**族谱通常由系谱树来表示。你的工作是统计那些没有孩子的家庭成员总数。**

# 输入格式： #

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.

**每个输入文件包含一个测试用例。每一种情况都以包含以下内容：第一行有一个正整数N（0<N<100），它代表节点的总数，还有一个正整数M，代表不是叶节点的节点总数；然后接着有M行，每一行的格式都是 ID k ID\[1\] ID\[2\]…ID\[K\]，其中第一个ID是一个两位数，代表一个给定的非叶节点，K是它的子节点的数量，后面是它的全部子节点的两位数ID序列。为了简单起见，规定根节点的ID为01。**

# 输出格式： #

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.

**对于每个测试用例，你应该从根节点开始，计算每一层（同辈）的没有孩子的家庭成员总数。结果必须在一行中输出，用空格隔开，并且在每行的末尾不能有额外的空格。示例示例表示一棵只有2个节点的树，其中01是根节点，02是唯一的子节点。因此在根01层上，有0个叶节点;下一层，有1个叶节点。所以应该在一行中输出0 1。**

# 输入样例： #

2 1
    01 1 02

# 输出样例： #

0 1

# 思路： #

使用`深度优先遍历`算法，但要注意，必须额外设置一个变量来弄清节点是否是叶节点。

## C实现代码： ##

#include <stdio.h>
    
    int N,M,a,b,map[101][101],book[101],num[101];
    //节点数、非叶节点数、父节点、子节点、存图数组、标记数组、统计数组
    const int max = 99999999;//最大值
    
    void dfs(int cur,int dep)//当前节点符号，当前深度
    { 
        int x = 0;//用来判别该节点是否是叶节点
        for(int i = 1;i <= N;i++)
            if(book[i] == 0 && map[cur][i] != max)//不是叶节点
            { 
                book[i] = 1;
                x = 1;
                dfs(i,dep + 1);
                book[i] = 0;
            }
        if(x == 0)//是叶节点
            num[dep] += 1;
        return;
    }
    
    int main()
    { 
        int c;//工具数
        for(int i = 1;i <= 100;i++)//不可走的路径设值为max
            for(int j = 1;j <= 100;j++)
                if(i != j)
                    map[i][j] = max;
        scanf("%d%d",&N,&M);
        for(int i = 1;i <= M;i++)//可走的路径设值为1
        { 
            scanf("%d%d",&a,&c);
            for(int j = 1;j <= c;j++)
            { 
                scanf("%d",&b);
                map[a][b] = 1;
                map[b][a] = 1;
            }
        }
        book[1] = 1;//从1号开始走所以将其标记
        dfs(1,1);//深度优先函数
        for(int i = 1;i <= 100;i++)//求出最大深度
            if(num[i] != 0)
                c = i;
        for(int i = 1;i <= c;i++)//输出结果
            if(i != c)
                printf("%d ",num[i]);
            else if(i == c)
                printf("%d",num[i]);
        return 0;
    }