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  • pat 1004 Counting Leaves

    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, the number of nodes in a tree, and M (<), 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
    
     题目大意:
    求每一层上的叶子结点个数。
    思路:
    用bfs
    代码:
    #include <iostream>
    #include <stdio.h>
    #include <string.h>
    #define MaxSize 105
    using namespace std;
    struct Node
    {
        int k;
        int leaf[MaxSize];
    };
    int layer;
    int leafCount[MaxSize];
    Node *nodes[MaxSize];
    void bfs()
    {
        int seq[MaxSize];
        int front = 0, end = 0;
        int k, l; //当前层和下一层的结点个数
        Node *t;
        layer = 0;
        seq[end] = 1;
        end = (end + 1) % MaxSize;
        k = 1;
        l = 0;
        while (front != end)
        {
            t = nodes[seq[front]];
            front = (front + 1) % MaxSize;
            if (t->k == 0)
            {
                leafCount[layer]++;
            }
            else
            {
                l += t->k;
                for (int i = 0; i < t->k; i++)
                {
                    seq[end] = t->leaf[i];
                    end = (end + 1) % MaxSize;
                }
            }
            k--;
            if (k == 0)
            {
                layer++;
                k = l;
                l = 0;
            }
        }
    }
    int main()
    {
        memset(leafCount, 0, sizeof(leafCount));
        for (int i = 0; i < MaxSize; i++)
        {
            nodes[i] = new Node();
            nodes[i]->k = 0;
        }
        int n, m, k;
        int num;
        cin >> n >> m;
        if (n == 0)
            return 0;
        Node *n1;
        for (int i = 0; i < m; i++)
        {
            cin >> num >> k;
            n1 = nodes[num];
            n1->k = k;
            for (int j = 0; j < n1->k; ++j)
            {
                cin >> n1->leaf[j];
            }
        }
        bfs();
        for (int i = 0; i < layer; i++)
        {
            if (i != 0)
                cout << " " << leafCount[i];
            else
                cout << leafCount[i];
        }
        return 0;
    }

    感想:

    一开始初始化指针数组为NULL,然后输入的时候新建,可能哪个结点输入的时候没建,导致两个样例过不去。dfs的代码写起来好少。

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  • 原文地址:https://www.cnblogs.com/stormax/p/10921902.html
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