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  • 7-21 Counting Leaves (30分)

    A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.

    Input Specification:

    The input consists of several test cases, each 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.

    For example, the first 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
    1 0
    7 4
    01 2 02 03
    06 1 07
    02 2 04 05
    03 1 06
    0 0
    
     

    Sample Output:

    0 1
    1
    0 0 2 1

    代码:
    #include <stdio.h>
    #include <stdlib.h>
    struct Node {
        int index;
        int c[101];
    }s[101];
    int leave[101];
    int mheight;
    void count(int id,int height) {
        if(mheight < height)mheight = height;
        if(s[id].index == 0) {
            leave[height] ++;
            return;
        }
        for(int i = 0;i < s[id].index;i ++)
            count(s[id].c[i],height + 1);
    }
    int main() {
        int n,m,k,id,d;
        while(scanf("%d%d",&n,&m) && n) {
            mheight = 0;
            for(int i = 0;i <= n;i ++) {
                leave[i] = s[i].index = 0;
            }
            for(int i = 0;i < m;i ++) {
                scanf("%d%d",&id,&k);
                s[id].index = 0;
                while(k --) {
                    scanf("%d",&d);
                    s[id].c[s[id].index ++] = d;
                }
            }
            count(1,0);
            printf("%d",leave[0]);
            for(int i = 1;i <= mheight;i ++)
                printf(" %d",leave[i]);
            putchar('
    ');
        }
    }
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  • 原文地址:https://www.cnblogs.com/8023spz/p/12260758.html
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