7-7 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<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.
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
//时间太晚了，快12点了，注释明天再写
#include
#include
#include
#include
#include
using namespace std;
vector tree[120];
int exist[120];
int level[120];
int maxx = 1;
void bfs()
{
queue q;
q.push(1);
maxx = 1;
while(!q.empty())
{
int x = q.front();
q.pop();
int len = tree[x].size();
for(int i = 0; i < len; i++)
{
int id = tree[x][i];
q.push(id);
level[id] = level[x] + 1;
if(maxx < level[id]) maxx = level[id];
}
}
return;
}
int main()
{
int n,m;
int id;
int k;
int a;
int sum = 0;
while(scanf("%d %d",&n,&m)&&n)
{
fill(exist,exist+120,1);
fill(level,level+120,0);
level[1] = 1;
for(int i = 0; i < 120; i++)
tree[i].clear();
for(int i = 1; i <= m; i++)
{
cin>>id>>k;
exist[id] = 0;
for(int j = 0; j < k; j++)
{
cin>>a;
tree[id].push_back(a);
}
}
bfs();
int flag = 1;
for(int i = 1; i <= maxx; i++)
{
sum = 0;
for(int j = 1; j <= n; j++)
{
if(level[j] == i&& exist[j] == 1)
sum++;
}
if(flag)
{
printf("%d",sum);
flag = 0;
}
else
printf(" %d",sum);
}
cout<<endl;
}
return 0;
}
