HDOJ-1800 Flying to the Mars（贪心 + 字典树 + BST）

Flying to the Mars

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6271    Accepted Submission(s): 2053

Problem Description

In the year 8888, the Earth is ruled by the PPF Empire . As the population growing , PPF needs to find more land for the newborns . Finally , PPF decides to attack Kscinow who ruling the Mars . Here the problem comes! How can the soldiers reach the Mars ? PPF convokes his soldiers and asks for their suggestions . “Rush … ” one soldier answers. “Shut up ! Do I have to remind you that there isn’t any road to the Mars from here!” PPF replies. “Fly !” another answers. PPF smiles :“Clever guy ! Although we haven’t got wings , I can buy some magic broomsticks from HARRY POTTER to help you .” Now , it’s time to learn to fly on a broomstick ! we assume that one soldier has one level number indicating his degree. The soldier who has a higher level could teach the lower , that is to say the former’s level > the latter’s . But the lower can’t teach the higher. One soldier can have only one teacher at most , certainly , having no teacher is also legal. Similarly one soldier can have only one student at most while having no student is also possible. Teacher can teach his student on the same broomstick .Certainly , all the soldier must have practiced on the broomstick before they fly to the Mars! Magic broomstick is expensive !So , can you help PPF to calculate the minimum number of the broomstick needed .
For example :
There are 5 soldiers (A B C D E)with level numbers : 2 4 5 6 4;
One method :
C could teach B; B could teach A; So , A B C are eligible to study on the same broomstick.
D could teach E;So D E are eligible to study on the same broomstick;
Using this method , we need 2 broomsticks.
Another method:
D could teach A; So A D are eligible to study on the same broomstick.
C could teach B; So B C are eligible to study on the same broomstick.
E with no teacher or student are eligible to study on one broomstick.
Using the method ,we need 3 broomsticks.
……

After checking up all possible method, we found that 2 is the minimum number of broomsticks needed.

 

 

Input

Input file contains multiple test cases.
In a test case,the first line contains a single positive number N indicating the number of soldiers.(0<=N<=3000)
Next N lines :There is only one nonnegative integer on each line , indicating the level number for each soldier.( less than 30 digits);

 

Output

For each case, output the minimum number of broomsticks on a single line.

 

Sample Input

4 10 20 30 04 5 2 3 4 3 4

 

Sample Output

1 2

 

  1 /* 

  2    功能Function Description:     HDOJ 1800 

  3    开发环境Environment:          DEV C++ 4.9.9.1

  4    技术特点Technique:

  5    版本Version:

  6    作者Author:                   可笑痴狂

  7    日期Date:                      20120810

  8    备注Notes:

  9    题目意思：

 10         有n个士兵每个人有一个水平值，水平高的的人可以教低的人，意思就是求最少的扫帚，

 11         那么我们只要知道找到最大重复元素的次数即可，因为相同的人肯定不能共用一个，

 12         所以求得最少即为最大的重复次数

 13    注意：前置的0必须要去掉，例如数据

 14            3

 15            0

 16            00

 17            000

 18    输出    3

 19 

 20   这题和拦截导弹那题(动态规划求最长不降子序列)的区别是 导弹来的顺序是固定的不能改变，而这道题是可以随意变动的

 21 */

 22 

 23 /*

 24 //代码一：-------直接排序找最长平台(贪心算法)

 25 #include<cstdio>

 26 #include<cstring>

 27 #include<algorithm>

 28 using namespace std;

 29 

 30 int main()

 31 {

 32     int n,i,max,t;

 33     int a[3005];

 34     while(scanf("%d",&n)!=EOF)

 35     {

 36         for(i=0;i<n;++i)

 37             scanf("%d",&a[i]);

 38         sort(a,a+n);

 39         max=1;

 40         t=1;

 41         for(i=1;i<n;++i)

 42         {

 43             if(a[i]!=a[i-1])

 44                 t=1;

 45             else 

 46                 ++t;

 47             if(t>max)

 48                 max=t;

 49         }

 50         printf("%d\n",max);

 51     }

 52     return 0;

 53 }

 54 

 55 

 56 //代码二：---------字典树

 57 #include<cstdio>

 58 #include<cstring>

 59 #include<stdlib.h>

 60 

 61 struct node

 62 {

 63     int sum;

 64     node *next[26];    

 65 };

 66 node memory[1000000];

 67 int k;

 68 int max;

 69 

 70 void insert(char *t,node *T)

 71 {

 72     int i,id,j;

 73     node *p,*q;

 74     p=T;

 75     i=0;

 76     while(t[i])

 77     {

 78         id=t[i]-'0';

 79         if(p->next[id]==NULL)

 80         {

 81         //    q=(node *)malloc(sizeof(node));

 82             q=&memory[k++];

 83             q->sum=0;

 84             for(j=0;j<26;++j)

 85                 q->next[j]=NULL;

 86             p->next[id]=q;

 87         }

 88         p=p->next[id];

 89         ++i;

 90     }

 91     p->sum++;

 92     if(max<p->sum)

 93         max=p->sum;

 94 }

 95 

 96 int main()

 97 {

 98     int n,i;

 99     char s[35];

100     node *T;

101 //    T=(node *)malloc(sizeof(node));

102     T=&memory[0];

103     while(scanf("%d",&n)!=EOF)

104     {

105         max=0;

106         k=1;

107         T->sum=0;

108         for(i=0;i<26;++i)

109             T->next[i]=NULL;

110         while(n--)

111         {

112             scanf("%s",s);

113             i=0;

114             while(s[i++]=='0');

115             insert(&s[i-1],T);

116         }

117         printf("%d\n",max);

118     }

119     return 0;

120 }

121 

122 */

123 

124 //代码三：

125 

126 //BST代码：----二分查找树/二叉排序树

127 

128 #include <stdio.h>

129 #include <malloc.h>

130 struct node 

131 {

132     int num;

133     node *lc;

134     node *rc;

135     int time;

136 };

137 int max;

138 struct node *BST(struct node *root,int num)

139 {

140     if (root==NULL)       //根节点为空 建立根节点

141     {

142         root=(struct node *)malloc(sizeof(struct node));

143         root->lc=NULL;

144         root->rc=NULL;

145         root->num=num;

146         root->time=1;

147     }

148     else 

149     {

150         if (root->num==num)        // 根的name 和 s相同，次数+1

151         {

152              root->time++;

153              if (root->time>max) 

154                  max=root->time;

155         }

156         else if (root->num>num)    // 根比s大，找左子树

157             root->lc=BST(root->lc,num);

158         else                      // 根比s小，找右子树

159             root->rc=BST(root->rc,num);

160     }

161     return root;

162 } 

163 int main()

164 {

165     int n,i,num;

166     while (scanf("%d",&n)!=EOF)

167     {

168          struct node *root=NULL;

169          max=1;

170          for (i=0;i<n;i++)

171          {

172             scanf("%d",&num);

173             root=BST(root,num);

174          }

175          printf ("%d\n",max);

176     }

177     return 0;

178 }