1004. Counting Leaves (30)

 

时间限制
400 ms
内存限制
32000 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.



  1 #include <iostream>
  2 
  3 using namespace std;
  4 
  5  
  6 
  7 struct fun
  8 
  9 {
 10 
 11    int father;
 12 
 13    bool ifdo;
 14 
 15    int h;
 16 
 17 };
 18 
 19  
 20 
 21 fun F[101];
 22 
 23  
 24 
 25 int main()
 26 
 27 {
 28 
 29  
 30 
 31    int n;
 32 
 33    while(cin>>n)
 34 
 35    {
 36 
 37    int i;
 38 
 39         for(i=1;i<=n;i++)
 40 
 41 {
 42 
 43     F[i].ifdo=true;
 44 
 45 F[i].h=1;
 46 
 47 F[i].father=-1;
 48 
 49 }
 50 
 51  
 52 
 53 int m;
 54 
 55 cin>>m;
 56 
 57         int high=0;
 58 
 59         if(m==0)  cout<<1<<endl;
 60 
 61         else
 62 
 63 {
 64 
 65  
 66 
 67  int name1,name2,num;
 68 
 69        while(m--)
 70 
 71 {
 72 
 73       cin>>name1>>num;
 74 
 75  
 76 
 77   F[name1].ifdo=false;
 78 
 79       for(i=0;i<num;i++)
 80 
 81   {
 82 
 83      cin>>name2;
 84 
 85                      F[name2].father=name1;
 86 
 87   }
 88 
 89   
 90 
 91 }
 92 
 93  
 94 
 95               
 96 
 97              /*这题目陷阱在这，每行第一个ID并不是
 98 
 99    从小到大按循序排列的，所以不能在上个循环中直接求h
100 
101    需要记录父节点   在以下循环中求h；
102 
103 */
104 
105  
106 
107  
108 
109 for(i=2;i<=n;i++)
110 
111  {
112 
113     F[i].h=F[F[i].father].h+1;
114 
115 if(F[i].h>high) high=F[i].h;
116 
117  }
118 
119  
120 
121  
122 
123  int j;
124 
125              bool fir=true;
126 
127          for(i=1;i<=high;i++)
128 
129 {
130 
131         int sum=0;
132 
133        for(j=1;j<=n;j++)
134 
135    {
136 
137           if(F[j].ifdo&&F[j].h==i) sum++;
138 
139    }
140 
141                if(fir) cout<<sum;
142 
143         else  cout<<" "<<sum;
144 
145         fir=false;
146 
147 }
148 
149           cout<<endl;
150 
151 }
152 
153  
154 
155         
156 
157    }
158 
159    
160 
161  
162 
163    return 0;
164 
165 } 
166 
167 
168
View Code

 

posted @ 2015-01-31 07:33  小爷  阅读(163)  评论(0)  编辑  收藏  举报