poj 3294 Life Forms

Life Forms
Time Limit: 5000MS   Memory Limit: 65536K
Total Submissions: 18777   Accepted: 5530

Description

You may have wondered why most extraterrestrial life forms resemble humans, differing by superficial traits such as height, colour, wrinkles, ears, eyebrows and the like. A few bear no human resemblance; these typically have geometric or amorphous shapes like cubes, oil slicks or clouds of dust.

The answer is given in the 146th episode of Star Trek - The Next Generation, titled The Chase. It turns out that in the vast majority of the quadrant's life forms ended up with a large fragment of common DNA.

Given the DNA sequences of several life forms represented as strings of letters, you are to find the longest substring that is shared by more than half of them.

Input

Standard input contains several test cases. Each test case begins with 1 ≤ n ≤ 100, the number of life forms. n lines follow; each contains a string of lower case letters representing the DNA sequence of a life form. Each DNA sequence contains at least one and not more than 1000 letters. A line containing 0 follows the last test case.

Output

For each test case, output the longest string or strings shared by more than half of the life forms. If there are many, output all of them in alphabetical order. If there is no solution with at least one letter, output "?". Leave an empty line between test cases.

Sample Input

3
abcdefg
bcdefgh
cdefghi
3
xxx
yyy
zzz
0

Sample Output

bcdefg
cdefgh

?

Source

Waterloo Local Contest, 2006.9.30
 
给定n个字符串，求出现在超过n/2个字符串中的最长子串。 
将n个字符串连起来，中间用不相同的且没有出现在字符串中的字符隔开， 求后缀数组。
然后二分答案，将后缀分成若干组，判断每组的后缀是否出现在不小于k个的原串中。这个做法的时间复杂度为O(nlogn)。
 
  1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdio>
  4 #include<cmath>
  5 #include<algorithm> 
  6 using namespace std;  
  7 int const N=100000+1000;  
  8 int sa[N],wa[N<<1],wb[N<<1],rk[N],num[N],h[N],wv[N],vis[101],id[N],s[N];   
  9 char s1[1003]; 
 10 int cmp(int *r,int x,int y,int z){
 11     return r[x]==r[y] && r[x+z]==r[y+z];  
 12 }
 13 void build_sa(int *r,int *sa,int n,int m){ 
 14     int i,j,p,*x=wa,*y=wb;  
 15     for(i=0;i<m;i++) num[i]=0; 
 16     for(i=0;i<n;i++) num[x[i]=r[i]]++; 
 17     for(i=1;i<m;i++) num[i]+=num[i-1];  
 18     for(i=n-1;i>=0;i--) sa[--num[x[i]]]=i;
 19     for(j=1,p=1;p<n;j<<=1,m=p){
 20         for(p=0,i=n-j;i<n;i++) y[p++]=i;  
 21         for(i=0;i<n;i++) if(sa[i]>=j) y[p++]=sa[i]-j;  
 22         for(i=0;i<m;i++) num[i]=0; 
 23         for(i=0;i<n;i++) wv[i]=x[y[i]]; 
 24         for(i=0;i<n;i++) num[wv[i]]++;  
 25         for(i=1;i<m;i++) num[i]+=num[i-1];  
 26         for(i=n-1;i>=0;i--) sa[--num[wv[i]]]=y[i];  
 27         swap(x,y);  
 28         for(i=1,p=1,x[sa[0]]=0;i<n;i++) 
 29             x[sa[i]]=cmp(y,sa[i],sa[i-1],j)? p-1: p++;  
 30     }
 31     for(i=0;i<n;i++) rk[i]=x[i]; 
 32 }
 33 void build_h(int *r,int *sa,int n){
 34     int k=0;  
 35     for(int i=0;i<n;i++){
 36         if(k) k--; 
 37         int j=sa[rk[i]-1]; 
 38         while (r[i+k]==r[j+k]) k++;  
 39         h[rk[i]]=k; 
 40     }
 41 }
 42 int check(int mid,int n,int tot){
 43     if(tot==1) return 1;  
 44     int num=0; 
 45     memset(vis,0,sizeof(vis));  
 46     for(int i=1;i<=n;i++){
 47         if(h[i]>=mid){
 48             if(!vis[id[sa[i-1]]]){
 49                 num++; vis[id[sa[i-1]]]=1; 
 50             }
 51             if(!vis[id[sa[i]]]){
 52                 num++; vis[id[sa[i]]]=1; 
 53             }
 54             if(num>tot/2) return 1; 
 55         }else {
 56             num=0;  
 57             memset(vis,0,sizeof(vis));  
 58         }
 59     }
 60     return 0; 
 61 }
 62 int solve(int mid,int n,int tot){
 63     int num=0; 
 64     memset(vis,0,sizeof(vis));  
 65     for(int i=1;i<=n;i++){
 66         if(h[i]>=mid){
 67             if(!vis[id[sa[i-1]]]){
 68                 num++; vis[id[sa[i-1]]]=1; 
 69             }
 70             if(!vis[id[sa[i]]]){
 71                 num++; vis[id[sa[i]]]=1; 
 72             }
 73         }else {
 74             if(num>tot/2) {
 75                 for(int j=sa[i-1];j<sa[i-1]+mid;j++) printf("%c",s[j]); 
 76                 printf("\n");
 77             }
 78             num=0;  
 79             memset(vis,0,sizeof(vis));  
 80         }
 81     }
 82     if(num>tot/2) {
 83         for(int j=sa[n];j<sa[n]+mid;j++) printf("%c",s[j]); 
 84         printf("\n");
 85     }
 86 }    
 87 int main(){
 88     int n; int t=0;  
 89     while (scanf("%d",&n)!=EOF && n){
 90         if(t++) printf("\n");  
 91         memset(s,0,sizeof(s));  
 92         int sum=130,len=0;  
 93         for(int i=0;i<n;i++){
 94             scanf("%s",s1);  
 95             int l=len;  
 96             int k=strlen(s1);  
 97             len+=k+1;  
 98             for(int j=l;j<len-1;j++) id[j]=i+1,s[j]=s1[j-l];  
 99             s[len-1]=sum++;  
100         }  
101         build_sa(s,sa,len+1,sum);   
102         build_h(s,sa,len); 
103         int l=0,r=1000,ans=0;  
104         while (l<=r){
105             int mid=(l+r)/2; 
106             if(check(mid,len,n)) {
107                 ans=mid;  
108                 l=mid+1; 
109             }else r=mid-1;  
110         }
111         if(ans==0) printf("?\n"); 
112         else solve(ans,len,n);  
113     }
114     return 0; 
115 }

 

posted @ 2019-06-03 13:54  zjxxcn  阅读(201)  评论(0)  编辑  收藏  举报