PAT (Advanced Level) 1004 Counting Leaves (30 分)

PAT (Advanced Level) 1004 Counting Leaves (30 分)

序：
又是一道30分的题，仔细一看，这是一道层次遍历多叉树的问题，用queue应该可以搞定。

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

题目概述：
给出一棵树的节点数，以及所有爸爸节点及其的儿子节点。求每一层中的叶子节点。

分析：
1.找根，根节点是唯一一个不在儿子节点的节点，用一个数组进行0，1记录
2.存储，这里我用了一个map，把每个爸爸和对应的儿子序列对应
3.遍历，用queue遍历，一层层遍历。若遇到节点无儿子，cnt++；否则，将其儿子push进queue，等待下一层遍历

代码如下：

//层次遍历&队列
//找根
#include<bits/stdc++.h>
using namespace std;int visit[110];
unordered_map<int,vector<int>> mp;int main()
{int n, m;scanf("%d%d", &n, &m);int father, k, temp;for(int i = 0; i < m; i++){scanf("%d%d", &father, &k);for(int j = 0; j < k; j++){scanf("%d", &temp);visit[temp] = 1;mp[father].push_back(temp);}}int root = 1;while(visit[root] != 0) root++;queue<int> qq;qq.push(root);int floor = 1;while(!qq.empty()){int len = qq.size();int cnt = 0, ft;for(int i = 0; i < len; i++){ft = qq.front();qq.pop();if(mp[ft].size() == 0) cnt++;else{for(int j = 0; j < mp[ft].size(); j++)qq.push(mp[ft][j]);}}if(floor == 1) printf("%d", cnt);else printf(" %d", cnt);floor++;}printf("n");return 0;
}

总结：
1.queue中的front()是取队首元素，pop()踢掉队首元素
2.用floor作为表示，区分输出的空格
3.root应该从1开始遍历，因为题目中的节点序号对应
4.题目中说01固定为根，但我们给出的是通解。我们的解法可以适用不定根的情况。

2022.01.08 简单bfs就好 以前花里胡哨的

//bfs罢了
#include<bits/stdc++.h>
using namespace std;vector<vector<int>> v;int main() {int n, m;cin >> n >> size(n + 1);for(int i = 0; i < m; i++) {int id, k;cin >> id >> k;for(int j = 0; j < k; j++) {int child;cin >> child;v[id].push_back(child);}}//开始bfsint level = 0;queue<int> q;q.push(1);int flag = 0;while(!q.empty()) {int countLeaves = 0;int len = q.size();for(int i = 0; i < len; i++) {int now = q.front();q.pop();if(v[now].size() == 0) ++countLeaves;else {for(int j = 0; j < v[now].size(); j++) q.push(v[now][j]);}}if(flag == 0) {++flag;cout << countLeaves;}else {cout << " " << countLeaves;}}return 0;
}