1004 Counting Leaves

最新推荐文章于 2022-07-26 16:28:25 发布

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本文介绍了一种算法，该算法通过输入家族树结构来计算每个层级中没有子节点的家庭成员数量。采用层次遍历的方法，首先标记非叶子节点，然后逐层计算并输出每个层级的叶子节点数量。

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
解题思路：在输入的时候就标记哪些节点是非叶子节点，然后层次遍历就好

#include<iostream>
#include<stdio.h>
#include<vector>
using namespace std;
struct node{
    int size;
    vector<int>children;
};
int main(){
    for (int n, k; scanf("%d %d", &n, &k) != EOF;){
        node arr[105];
        //记录节点是否是叶子节点
        int flag[105] = {0};
        for (int i = 0; i < k; i++){
            int temp,size;
            scanf("%d %d", &temp,&size);
            arr[temp].size = size;
            flag[temp] = 1;
            while (size--){
                int child;
                scanf("%d", &child);
                arr[temp].children.push_back(child);
            }
        }
        int count = 1;
        int every_count = 0;
        if (flag[1]){
            printf("0");
        }
        else{
            printf("1");
        }
        vector<node>vec;
        for (int i = 0; i < arr[1].size; i++){
            if (!flag[arr[1].children[i]]){
                every_count++;
            }
            vec.push_back(arr[arr[1].children[i]]);
            count++;
        }
        if (vec.size()){
            printf(" %d", every_count);
        }
        while (count != n){
            vector<node>haha;
            every_count = 0;
            for (int i = 0; i < vec.size(); i++){
                for (int j = 0; j < vec[i].size; j++){
                    if (!flag[vec[i].children[j]]){
                        every_count++;
                    }
                    haha.push_back(arr[vec[i].children[j]]);
                    count++;
                }
            }
            vec = vector<node>(haha);
            printf(" %d", every_count);
        }
        printf("\n");
    }
    return 0;
}