| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 8683 | Accepted: 2375 |
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 ?
1 #include <iostream> 2 #include <stdio.h> 3 #include <math.h> 4 #include <vector> 5 #include <string.h> 6 using namespace std; 7 #define N 101000 8 vector<int> ai; 9 int a[N],c[N],d[N],e[N],sa[N],height[N],n,b[N],m,t,jilu[110],bi[110]; 10 int cmp(int *r,int a,int b,int l) 11 { 12 return r[a]==r[b]&&r[a+l]==r[b+l]; 13 } 14 void da() 15 { 16 int i,j,p,*x=c,*y=d,*t; 17 memset(b,0,sizeof(b)); 18 for(i=0; i<n; i++)b[x[i]=a[i]]++; 19 for(i=1; i<m; i++)b[i]+=b[i-1]; 20 for(i=n-1; i>=0; i--)sa[--b[x[i]]]=i; 21 for(j=1,p=1; p<n; j*=2,m=p) 22 { 23 for(p=0,i=n-j; i<n; i++)y[p++]=i; 24 for(i=0; i<n; i++)if(sa[i]>=j)y[p++]=sa[i]-j; 25 for(i=0; i<n; i++)e[i]=x[y[i]]; 26 for(i=0; i<m; i++)b[i]=0; 27 for(i=0; i<n; i++)b[e[i]]++; 28 for(i=1; i<m; i++)b[i]+=b[i-1]; 29 for(i=n-1; i>=0; i--)sa[--b[e[i]]]=y[i]; 30 for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1; i<n; i++) 31 x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++; 32 } 33 } 34 void callheight() 35 { 36 int i,j,k=0; 37 b[0]=0; 38 for(i=1; i<n; i++)b[sa[i]]=i; 39 for(i=0; i<n-1; height[b[i++]]=k) 40 for(k?k--:0,j=sa[b[i]-1]; a[i+k]==a[j+k]; k++); 41 } 42 int fun(int x) 43 { 44 int i; 45 for(i=0; i<t; i++) 46 { 47 if(jilu[i]>=x)break; 48 } 49 return i; 50 } 51 bool check(int mid,int y) 52 { 53 int i=0; 54 while(1) 55 { 56 while(i<n&&height[i]<mid)i++; 57 if(i==n)break; 58 memset(bi,0,sizeof(bi)); 59 int u=fun(sa[i-1]),uu=sa[i-1]; 60 bi[u]=1; 61 int sum=1; 62 while(i<n&&height[i]>=mid) 63 { 64 int yu=fun(sa[i]); 65 if(yu==u) 66 { 67 i++; 68 continue; 69 } 70 else if(!bi[yu]) 71 { 72 bi[yu]=1; 73 sum++; 74 u=yu; 75 } 76 i++; 77 } 78 if(sum>t/2) 79 { 80 if(y) 81 ai.push_back(uu); 82 else 83 return 1; 84 } 85 } 86 return 0; 87 } 88 void check1(int mid) 89 { 90 int i=1; 91 while(1) 92 { 93 while(i<n&&height[i]<mid)i++; 94 if(i==n)break; 95 memset(bi,0,sizeof(bi)); 96 int u=fun(sa[i-1]),uu=sa[i-1]; 97 bi[u]=1; 98 int sum=1; 99 while(i<n&&height[i]>=mid) 100 { 101 int yu=fun(sa[i]); 102 if(yu==u) 103 { 104 i++; 105 continue; 106 } 107 else if(!bi[yu]) 108 { 109 bi[yu]=1; 110 sum++; 111 u=yu; 112 } 113 i++; 114 } 115 if(sum>t/2)ai.push_back(uu); 116 } 117 } 118 int main() 119 { 120 int i,j,temp,ll=0; 121 char x; 122 while(scanf("%d",&t)&&t) 123 { 124 ai.clear(); 125 if(ll)printf(" "); 126 ll++; 127 n=0; 128 temp=30; 129 x=getchar(); 130 for(i=0; i<t; i++) 131 { 132 while(x=getchar()) 133 { 134 if(x==' ')break; 135 a[n++]=x-'a'+1; 136 } 137 jilu[i]=n-1; 138 a[n++]=temp++; 139 } 140 m=150; 141 a[n-1]=0; 142 da(); 143 callheight(); 144 int l=0,r=1001; 145 while(l<=r) 146 { 147 int mid=(l+r)>>1; 148 if(check(mid,0)) 149 l=mid+1; 150 else r=mid-1; 151 } 152 if(r) 153 { 154 check(r,1); 155 for(j=0; j<ai.size(); j++) 156 { 157 for(i=0; i<r; i++) 158 { 159 putchar('a'+a[ai[j]+i]-1); 160 } 161 printf(" "); 162 } 163 } 164 else 165 { 166 printf("? "); 167 } 168 } 169 }