zoukankan      html  css  js  c++  java
  • 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 levelstarting 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算法 
    #include<cstdio>
    #include<vector>
    using namespace std;
    const int maxn = 110;
    vector<int> node[maxn];
    int hashtable[maxn] = {0}; 
    int maxlevel = 0;
    void DFS(int index,int level){
        if(node[index].size() == 0){
            hashtable[level]++;
            if(level > maxlevel) maxlevel = level;
            return;
        }
        for(int i = 0; i < node[index].size(); i++){
            DFS(node[index][i],level+1);
        }
    }
    
    int main(){
       int n,m,father,child,k;
       scanf("%d%d",&n,&m);
       for(int i = 0; i < m; i++){
           scanf("%d%d",&father,&k);
           for(int j = 0; j < k; j++){
               scanf("%d",&child);
               node[father].push_back(child);
           }
       }
       DFS(1,0);
       for(int i = 0; i <= maxlevel; i++){
           if(i == 0)  printf("%d",hashtable[i]);
           else printf(" %d",hashtable[i]);
           }
           return 0;
    } 
    #include<cstdio>
    #include<queue>
    #include<vector>
    #include<algorithm>
    using namespace std;
    const int maxn = 110;
    
    vector<int> G[maxn];
    int h[maxn] = {0}; //树高度
    int leaf[maxn] = {0}; //每层节点统计
    int max_h = 0;
    void BFS(){
        queue<int> q;
        q.push(1);
        while(!q.empty()){
            int id = q.front();
            q.pop();
            max_h = max(max_h,h[id]);
            if(G[id].size() == 0){
                leaf[h[id]]++;
            }
            for(int i = 0; i < G[id].size(); i++){
                h[G[id][i]] = h[id] + 1;
                q.push(G[id][i]);
            }
        }
    }
    
    int main(){
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i = 0; i < m; i++){
            int father,child,k;
            scanf("%d%d",&father,&k);
            for(int j = 0; j < k; j++){
                scanf("%d",&child);
                G[father].push_back(child);
            }
        }
        h[1] = 1;
        BFS();
        for(int i = 1; i <= max_h; i++){
            if(i == 1)printf("%d",leaf[i]);
            else printf(" %d",leaf[i]);
        }
    
        return 0;
    } 
  • 相关阅读:
    字符串初始化、查找字符+获取字符
    冒泡排序
    JAVA中值类型和引用类型的不同?
    二维数组初始化,属性,遍历,输出各元素总和。
    数组定义属性遍历循环,输出最大数
    for穷举,叠代练习
    HTML--Boby部分之<a>标签
    HTML--Boby内标签之多行文本及下拉框
    HTML--Boby部分Input之重置
    HTML--Boby部分Input之上传文件
  • 原文地址:https://www.cnblogs.com/wanghao-boke/p/9295457.html
Copyright © 2011-2022 走看看