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 0Sample Output
1 2 0 1 0 0 0 0 0Hint
Huge input file, 'scanf' recommended to avoid TLE.Source
Central Europe 2005
#include<cstdio> #include<cstring> struct dictree { int child[4];/*假设只有26个字符*/ int n;/*记录当前单词出现的次数*/ } tree[400010]; int a[20010]; int root; int nownode=0; int n,m; int t[30]; void insert(char *source) { int i,j; int current,newnode; current=root;//从根节点开始寻找 for(i=0; i<m; i++)/*逐个插入字符*/ { int tt=t[source[i]-'A']; if(tree[current].child[tt]!=-1)/*存在*/ { current=tree[current].child[tt]; } else { ++nownode;//看现在第几个node还未被使用 newnode=nownode; for(j=0; j<4; j++) tree[newnode].child[j]=-1;//初始化 tree[current].child[tt]=newnode; current=newnode; tree[current].n=0; } } tree[current].n++; } int main() { char temp[21]; int i; t[0]=0; t[2]=1; t[6]=2; t[19]=3; while(scanf("%d%d",&n,&m)!=EOF) { if(n==0&&m==0) break; memset(a,0,sizeof(a)); root=0; nownode=0; for(i=0; i<4; i++) tree[root].child[i]=-1;//初始化 tree[root].n=0; for(int i=1;i<=n;i++) { scanf("%s",temp); insert(temp); } for(int i=0;i<=nownode;i++) if(tree[i].n!=0) a[tree[i].n]++; for(int i=1;i<=n;i++) printf("%d/n",a[i]); } return 0; }
