Life Forms

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 7270		Accepted: 1998

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

3abcdefgbcdefghcdefghi3xxxyyyzzz0

Sample Output

bcdefgcdefgh?

Source

, 2006.9.30

 

 

依旧二分判定，只是要先找出最大长度，然后在根据这个长度再判定输出。对于长度相同的，用tag标记以免重复。

 

#include
     
      //最长公共子串#include
      
       #define Max(a, b)   a>b?a:bconst int maxn = 101001 ;int wa[maxn], wb[maxn], wv[maxn], ws[maxn], rank[maxn], height[maxn], sa[maxn], s[maxn], loc[maxn] ;bool vis[1001] ;char str[1001], ans[1001] ;int n ;int cmp(int *r,int a,int b,int l){    return r[a]==r[b] && r[a+l]==r[b+l] ;}int abs(int a){    return a<0?-a:a ;}void da(int *r, int n, int m){    int i, j, p, *x = wa, *y = wb, *t ;    for(i=0; i
       
        =0; i--) sa[--ws[x[i]]] = i ;    for(j=1,p=1; p
        
         =j) y[p++] = sa[i] - j ;        for(i=0; i
         
          =0; i--) sa[--ws[wv[i]]] = y[i] ;        for(t=x, x=y, y=t, p=1, x[sa[0]]=0, i=1; i
          
           n/2) return 1 ; } return 0 ;}void print(int mid, int len){ int count = 0, tag = 0 ; memset(vis, false, sizeof(vis)) ; for(int i=2; i<=len; i++){ if(height[i]
           
            n/2&&!tag){ for(int j=0; j
            
             =left){ mid = (right + left) / 2 ; if(check(mid, m)){ left = mid + 1 ; flag = mid ; } else right = mid - 1 ; } if(flag){ print(flag, m) ; printf("\n") ; } else printf("?\n\n") ; } return 0 ;}
            
           
          
         
        
       
      
     

 

 

下面我的代码 RE

 

#include
     
      #include
      
       const int maxn=111111;int wa[maxn],wb[maxn],wv[maxn],ws[maxn];int rank[maxn],height[maxn],sa[maxn],s[maxn],loc[maxn];bool vis[1010];char str[1010],ans[1010];int n;int cmp(int *r,int a,int b,int l){    return r[a]==r[b] && r[a+l]==r[b+l];}void da(int *r,int n,int m){    int i,j,p,*x=wa,*y=wb,*t;    for(i=0;i
       
        =0;i--) sa[--ws[x[i]]]=i;    for(j=1,p=1;p
        
         =j) y[p++]=sa[i]-j;        for(i=0;i
         
          =0;i--) sa[--ws[wv[i]]]=y[i];        for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1;i
          
           n/2) return 1; } return 0;}void print(int mid,int len){ int count=0,tag=0; memset(vis,0,sizeof(vis)); for(int i=2;i<=len;i++){ if(height[i]
           
            n/2 && !tag){ for(int j=0;j
            
             =left){ mid=(left+right)>>1; if(check(mid,m)){ left=mid+1; flag=mid; }else right=mid-1; } if(flag){ print(flag,m); printf("\n"); }else printf("?\n\n"); } return 0;}