1004 Counting Leaves（树，层次遍历）-蒲公英云

1004 Counting Leaves（树，层次遍历）

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;
}