1004. Counting Leaves (30)

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

  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.

Output

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

DFS解法

#include <iostream>
#include <cstdio>
#include <vector>
#include <queue>
using namespace std;
int n,m;
int* result;
int max_h;
struct Node{/*孩子兄弟表示法*/
    int firstChild=-1;
    vector<int> brother;
};
void dfs(Node* tree,int id,int deep){
    max_h=deep>max_h?deep:max_h;
    //判断当前的id是不是叶子节点
    if(tree[id].firstChild==-1)
        result[deep]++;
    for(int i=0;i<tree[id].brother.size();i++){
        dfs(tree,tree[id].brother[i],deep);
    }
    if(tree[id].firstChild!=-1)
        dfs(tree,tree[id].firstChild,deep+1);
}
int main()
{
    //freopen("in.txt","r",stdin);
    cin>>n>>m;
    Node* tree=new Node[100+10];
    result=new int[n]{0};
    for(int i=0;i<m;i++){
        int id,k,id_1;
        cin>>id>>k;
        cin>>id_1;
        tree[id].firstChild=id_1;
        for(int j=1;j<k;j++){
            int b;
            cin>>b;
            tree[id_1].brother.emplace_back(b);
        }
    }
    dfs(tree,1,0);
    for(int i=0;i<max_h;i++){
        cout<<result[i]<<" ";
    }
    cout<<result[max_h];
    return 0;
}

采用孩子兄弟来存储，也可以用邻接表存储。要注意，树的高度要用变量存储，不能直接认为是m

BFS解法

#include <iostream>
#include <cstdio>
#include <vector>
#include <queue>
using namespace std;
int n,m;
int* result;
int max_h;
int* h;//记录节点所在的层次
struct Node{/*孩子兄弟表示法*/
    int firstChild=-1;
    vector<int> brother;
};
/*void dfs(Node* tree,int id,int deep){
    max_h=deep>max_h?deep:max_h;
    //判断当前的id是不是叶子节点
    if(tree[id].firstChild==-1)
        result[deep]++;
    for(int i=0;i<tree[id].brother.size();i++){
        dfs(tree,tree[id].brother[i],deep);
    }
    if(tree[id].firstChild!=-1)
        dfs(tree,tree[id].firstChild,deep+1);
}*/
void bfs(Node* tree){
    max_h=0;
    queue<int> q;
    q.push(1);
    while(!q.empty()){
        int id=q.front();
        q.pop();
        max_h=h[id]>max_h?h[id]:max_h;
        if(tree[id].firstChild==-1){//是叶子节点
            result[h[id]]++;
            continue;
        }
        //更新当前节点所有孩子节点的层次
        h[tree[id].firstChild]=h[id]+1;
        q.push(tree[id].firstChild);
        for(int i:tree[tree[id].firstChild].brother){
            h[i]=h[id]+1;
            q.push(i);
        }
    }
}
int main()
{
    //freopen("in.txt","r",stdin);
    cin>>n>>m;
    Node* tree=new Node[100+10];
    result=new int[n]{0};
    h=new int[n+1]{0};
    for(int i=0;i<m;i++){
        int id,k,id_1;
        cin>>id>>k;
        cin>>id_1;
        tree[id].firstChild=id_1;
        for(int j=1;j<k;j++){
            int b;
            cin>>b;
            tree[id_1].brother.emplace_back(b);
        }
    }
    //dfs(tree,1,0);
    bfs(tree);
    for(int i=0;i<max_h;i++){
        cout<<result[i]<<" ";
    }
    cout<<result[max_h];
    return 0;
}