1004. Counting Leaves （树）

1004. Counting Leaves (30)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

2 101 1 02

Sample Output

0 1

解题思路：题目百度翻译看了好久终于看懂了。。。根据题目所给的节点建树，然后找树的每一层叶子（没有子结点的结点）的个数。。。直接深搜一遍就可以了，用一个数组来存每一层的叶子个数

代码如下：

#include <stdio.h>#include <math.h>#include <string.h>#include <algorithm>using namespace std;int b[101],a[101][101],s[101],maxx;void dfs(int n,int m){    if(b[n]==0)    {        s[m]++;        maxx = maxx>m ? maxx : m;///记录最大层数        return;    }    for(int i=0;i<b[n];i++)///深搜每一个子结点        dfs(a[n][i],m+1);}int main(){    int n,m,id,k;    scanf("%d %d",&n,&m);    while(m--)    {        scanf("%d %d",&id,&k);        b[id]=k;///存储id的子结点数        for(int i=0;i<k;i++)            scanf("%d",&a[id][i]);    }    dfs(1,0);    for(int i=0;i<maxx;i++)        printf("%d ",s[i]);    printf("%d\n",s[maxx]);    return 0;}