1004. Counting Leaves (30)

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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h100 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

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