1004 Counting Leaves（树，层次遍历）

秒速五厘米 2024-04-08 08:43 82阅读 0赞

## 1004 Counting Leaves ##

### 0、题目 ###

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

### 1、大致题意 ###

输出每层叶节点的数目

### 2、基本思路 ###

定义结点的结构体`Node`，采用`STL`模板的`queue`来进行层序遍历，遍历过程中将孩子结点的层数+1，leaves数组用来存储每层叶节点的数目

### 3、AC代码 ###

#include<cstdio>
    #include<queue>
    #include<vector>
    using namespace std;
    
    const int maxn = 110;
    
    typedef struct {
        
    	int layer;
    	vector<int> child;
    } Node;
    
    int leaves[maxn], maxLayer = -1;
    Node node[maxn];
    
    void levelOrder(int st) {
        
    	queue<int> q;
    	q.push(st);
    	node[st].layer = 1;
    	while(!q.empty()) {
        
    		int now = q.front();
    		q.pop();
    		if(node[now].child.empty()) {
        
    			leaves[node[now].layer]++; //这一层的叶结点数目+1
    			if(node[now].layer > maxLayer) {
         //最大层的节点一定为叶结点
    				maxLayer = node[now].layer;
    			}
    		}
    		for(int i = 0; i < node[now].child.size(); i++) {
        
    			q.push(node[now].child[i]);
    			node[node[now].child[i]].layer = node[now].layer + 1; //孩子结点的层数+1
    		}
    	}
    }
    
    int main(int argc, char** argv) {
        
    	int N, M;
    	scanf("%d %d", &N, &M);
    	for(int i = 0; i < M; i++) {
        
    		int father, K, tempChild;
    		scanf("%d %d", &father, &K);
    		for(int j = 0; j < K; j++) {
        
    			scanf("%d", &tempChild);
    			node[father].child.push_back(tempChild);
    		}
    	}
    	levelOrder(1);
    	for(int i = 1; i <= maxLayer; i++) {
        
    		if(i != 1) printf(" ");
    		printf("%d", leaves[i]);
    	}
    	return 0;
    }