PAT (Advanced Level) Practise - 1004. Counting Leaves (30)

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本文介绍了一个计算家族树中每层叶子节点数量的算法问题。通过输入节点和非叶节点的信息，该算法能够计算出每层中没有子节点的成员数量，并按层级顺序输出这些数量。

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 1
01 1 02

Sample Output

0 1

题意：

给你一棵树，让你求出每层中有几个节点没有子节点。

我没有用DFS来做，我直接模拟了。

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <map>
using namespace std;
#define LL long long
const int maxn = 111;
const int inf = 0x3f3f3f3f;
int mp[maxn][maxn];
int flag[maxn];
int cnt=0;
int main()
{
    int n,m;
    memset(flag,-1,sizeof flag);
    scanf("%d %d",&n,&m);
    for(int i=1;i<=m;i++){
        int fa,k,son;
        scanf("%d %d",&fa,&k);
        for(int i=1;i<=k;i++){
            scanf("%d",&son);
            mp[fa][son]=1;
            flag[fa]=1;
            if(flag[son]==-1){
                flag[son]=0;
            }
        }

    }
    int num[maxn];
    num[++cnt]=1;
    int fnum[maxn];
    int pp[maxn];
    int pcnt=0;
    int *r=num,*p=fnum;
    while(cnt!=0){
        int fcnt=0;
        int ans=0;
        for(int i=1;i<=cnt;i++){
            int m=1;
            int u=r[i];
            for(int j=1;j<=99;j++){
                if(mp[u][j]){
                    p[++fcnt]=j;
                    m=0;
                }
            }
            if(m==1)
                ans++;
        }
        cnt=fcnt;
        swap(r, p);
        pp[++pcnt]=ans;
    }
    for(int i=1;i<=pcnt;i++){
        if(i-1) printf(" ");
        printf("%d",pp[i]);
    }
    printf("\n");

}