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  

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