1004 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:
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.
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 01level, 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#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
int tree[109][109];
int root;
int N;
void BFS( ){
queue < int > q1;
q1.push( root ) ;
int cur ,flag = 0 ;
while( 1 ){
int cnt=0;
queue < int > q2;
while( !q1.empty() ){
cur = q1.front();
q1.pop();
int f = 0 ;
for( int i =1;i<=N;i++){
if( tree[cur][i] != 0 ){
f = 1;
q2.push(i);
}
}
if( f == 0 ){
cnt++;
}
}
if( flag == 0 ){
flag = 1;
}
else printf(" ");
printf("%d",cnt);
if( q2.empty() ){
printf("\n");
break;
}
while( !q2.empty() ){
int t = q2.front();
q2.pop();
q1.push(t);
}
}
}
// 用层序遍历，肯定就出来了。
// 怎么做 ？
int main(void){
int m;
while( scanf("%d",&N)!=EOF &&N ){
scanf("%d",&m);
memset( tree , 0 ,sizeof( tree) );
char str[3];
for( int i=1;i<=m;i++){
scanf("%s",str);
int id = ( str[0] - '0' ) * 10 + str[1] - '0';
root = id;
int k;
scanf("%d",&k);
for( int j=1;j<=k;j++){
scanf("%s",str);
id = ( str[0] - '0') *10 + str[1] -'0';
tree[root][id] = 1;
}
}
root = 1;
BFS( );
}
return 0;
}