Find the Clones

Find the Clones

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 6365		Accepted: 2375

Description

Doubleville, a small town in Texas, was attacked by the aliens. They have abducted some of the residents and taken them to the a spaceship orbiting around earth. After some (quite unpleasant) human experiments, the aliens cloned the victims, and released multiple copies of them back in Doubleville. So now it might happen that there are 6 identical person named Hugh F. Bumblebee: the original person and its 5 copies. The Federal Bureau of Unauthorized Cloning (FBUC) charged you with the task of determining how many copies were made from each person. To help you in your task, FBUC have collected a DNA sample from each person. All copies of the same person have the same DNA sequence, and different people have different sequences (we know that there are no identical twins in the town, this is not an issue).

Input

The input contains several blocks of test cases. Each case begins with a line containing two integers: the number 1 ≤ n ≤ 20000 people, and the length 1 ≤ m ≤ 20 of the DNA sequences. The next n lines contain the DNA sequences: each line contains a sequence of m characters, where each character is either `A', `C', `G' or `T'. 
The input is terminated by a block with n = m = 0 .

Output

For each test case, you have to output n lines, each line containing a single integer. The first line contains the number of different people that were not copied. The second line contains the number of people that were copied only once (i.e., there are two identical copies for each such person.) The third line contains the number of people that are present in three identical copies, and so on: the i -th line contains the number of persons that are present in i identical copies. For example, if there are 11 samples, one of them is from John Smith, and all the others are from copies of Joe Foobar, then you have to print `1' in the first andthe tenth lines, and `0' in all the other lines.

Sample Input

9 6
AAAAAA
ACACAC
GTTTTG
ACACAC
GTTTTG
ACACAC
ACACAC
TCCCCC
TCCCCC
0 0

Sample Output

1
2
0
1
0
0
0
0
0
 题目大意：输入两个数m,n分别代表基因片段的数目和每个基因片段的长度，输出结果为n个数，第i个数代表出现次数为i-1的基因片段的数量。
时间限制是5000MS，时间特别宽松，用map都能过。   map,sort，AC自动机，Trie树都可以过。
map方法 2900+MS；map法的优点：编程毫无难度，思路及其简单，能在最短的时间内AC这题，在比赛上用这种方法的优势最大。当然如果是卡时间的话，我们考虑用sort  qsort。
数据结构题的特点：代码量大，编程复杂度高，很锻炼代码能力和编程思想。
在考场上最好的算法就是能在最少的时间内得到ac.
time:2900+ms;

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <string>
 #include <vector>
 #include <stack>
 #include <queue>
 #include <set>
 #include <map>
 #include <iomanip>
 #include <cstdlib>
 using namespace std;
 const int INF=0x5fffffff;
 const int MS=;
 const double EXP=1e-;
 int num[MS];
 char str[MS][];
 struct cmp
 {
     bool operator()(const char *a,const char *b)const
     {
         return strcmp(a,b)<;
     }
 };
 map<char *,int,cmp> mp;
 int main()
 {
     int n,m;
     char *s;
     while(scanf("%d%d",&n,&m)==&&(n+m))
     {
         mp.clear();
         int j=;
         for(int i=;i<n;i++)
         {
             s=str[j++];//需要不同的地址
             scanf("%s",s);
             mp[s]++;
         }
         memset(num,,sizeof(num));
         for(map<char*,int,cmp>::iterator it=mp.begin();it!=mp.end();it++)
         {
             num[it->second-]++;
         }
         for(int i=;i<n;i++)
         {
             printf("%d\n",num[i]);
         }
     }
 
     return ;
 }

Trie 树

time:204ms

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <string>
 #include <vector>
 #include <stack>
 #include <queue>
 #include <set>
 #include <map>
 #include <iomanip>
 #include <cstdlib>
 using namespace std;
 const int INF=0x5fffffff;
 const int MS=;
 const double EXP=1e-;
 
 struct node
 {
     // int id;
     //bool have;
     int n;
     node * next[];
 }nodes[MS];   //注意这个大小  尽量大一点,避免访问非法内存
 
 node *root;
 int cnt;
 int t[];
 int num[MS/];
 
 node * add_node(int c)
 {
     node *p=&nodes[c];
     for(int i=;i<;i++)
         p->next[i]=NULL;
    // p->have=false;
     p->n=;
     return p;
 }
 
 void insert(char *str)
 {
     node *p=root,*q;
     int len=strlen(str);
     for(int i=;i<len;i++)
     {
         int id=t[str[i]-'A'];
         if(p->next[id]==NULL)
         {
             p->next[id]=add_node(cnt);
             cnt++;
         }
         p=p->next[id];
     }
     p->n++;
 }
 
 int main()
 {
     int n,m,i;
     t[]=;
     t[]=;
     t[]=;
     t[]=;
     char str[];
     while(scanf("%d%d",&n,&m)==&&(n+m))
     {
         cnt=;
         memset(num,,sizeof(num));
         root=add_node(cnt);
         cnt++;
         for(i=;i<n;i++)
         {
             scanf("%s",str);
             insert(str);
         }
         int sum=;
         for(i=;i<=cnt;i++)
         {
             if(nodes[i].n)
             {
                 num[nodes[i].n-]++;
             }
         }
         for(i=;i<n;i++)
             printf("%d\n",num[i]);
     }
     return ;
 }