PAT甲级 - 1004 Counting Leaves （30 分）

迷南。 2022-03-19 16:29 129阅读 0赞

题目链接：[（PAT甲级）1004 Counting Leaves （30 分）][PAT_1004 Counting Leaves _30]

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

题解：其实这题就是树的层次遍历，dfs求出结点的层数。一开始，边输入边根据父结点的层数来求该结点的层数，得了19分，这个思路没有考虑到父结点的层数也许此时还没算出来。毕竟，输入时没什么规律。所以我们先得把父子关系存起来，输入结束后再来计算层数。

#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <vector>
    #include <cmath>
    using namespace std;
    const int maxn=110;
    vector<int> v[maxn];
    int layer[maxn];
    int mx=-1;
    
    void dfs(int id,int lay){
    	mx=max(mx,lay);
    	if(v[id].size()==0){
    		layer[lay]++;
    		return ;	
    	} 
    	for(int i=0;i<v[id].size();i++){
    		int x=v[id][i];
    		dfs(x,lay+1);
    	}
    }
    
    int main(){
    	int n,m;
    	while(~scanf("%d%d",&n,&m)&&n){
    		memset(layer,0,sizeof(layer));
    		for(int i=1;i<=n;i++) v[i].clear();
    		int id,k,a;
    		
    		while(m--){
    			scanf("%d%d",&id,&k);
    			while(k--){
    				scanf("%d",&a);
    				v[id].push_back(a);
    			}
    		}
    		dfs(1,0);
    		printf("%d",layer[0]);
    		for(int i=1;i<=mx;i++){
    			printf(" %d",layer[i]);
    		}
    		printf("\n");
    	}
    	return 0;
    }

[PAT_1004 Counting Leaves _30]: https://pintia.cn/problem-sets/994805342720868352/problems/994805521431773184