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，求出如今不小于k/2个字符串中的最长子串。
将n个字符串连起来，中间用不同样的且没有出如今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，推断每组的后缀是否出如今不小于k个的原串中。
这个做法的时间复杂度为O(nlogn)。
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;
#define LS 2*i
#define RS 2*i+1
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 1000005
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EXP 1e-8
int wa[N],wb[N],wsf[N],wv[N],sa[N];
int rank[N],height[N],s[N],a[N];
//sa:字典序中排第i位的起始位置在str中第sa[i]
//rank:就是str第i个位置的后缀是在字典序排第几
//height：字典序排i和i-1的后缀的最长公共前缀
int cmp(int *r,int a,int b,int k)
{
return r[a]==r[b]&&r[a+k]==r[b+k];
}
void getsa(int *r,int *sa,int n,int m)//n要包括末尾加入的0
{
int i,j,p,*x=wa,*y=wb,*t;
for(i=0; i<m; i++) wsf[i]=0;
for(i=0; i<n; i++) wsf[x[i]=r[i]]++;
for(i=1; i<m; i++) wsf[i]+=wsf[i-1];
for(i=n-1; i>=0; i--) sa[--wsf[x[i]]]=i;
p=1;
j=1;
for(; p<n; j*=2,m=p)
{
for(p=0,i=n-j; i<n; i++) y[p++]=i;
for(i=0; i<n; i++) if(sa[i]>=j) y[p++]=sa[i]-j;
for(i=0; i<n; i++) wv[i]=x[y[i]];
for(i=0; i<m; i++) wsf[i]=0;
for(i=0; i<n; i++) wsf[wv[i]]++;
for(i=1; i<m; i++) wsf[i]+=wsf[i-1];
for(i=n-1; i>=0; i--) sa[--wsf[wv[i]]]=y[i];
t=x;
x=y;
y=t;
x[sa[0]]=0;
for(p=1,i=1; i<n; i++)
x[sa[i]]=cmp(y,sa[i-1],sa[i],j)? p-1:p++;
}
}
void getheight(int *r,int n)//n不保存最后的0
{
int i,j,k=0;
for(i=1; i<=n; i++) rank[sa[i]]=i;
for(i=0; i<n; i++)
{
if(k)
k--;
else
k=0;
j=sa[rank[i]-1];
while(r[i+k]==r[j+k])
k++;
height[rank[i]]=k;
}
}
char str[N];
int len[105],size,ans[N];
bool vis[105];
int check(int mid,int n,int k)
{
int i,j;
int size = 0,cnt = 0;
MEM(vis,false);
for(i = 1; i<=n; i++)
{
if(height[i]>=mid)
{
for(j = 1; j<=k; j++)
{
//把sa[i-1]或sa[i]所在的字符串给标记。相同的串不反复累加
if(sa[i]>len[j-1]&&sa[i]<len[j]) cnt+=(vis[j]?0:1),vis[j]=true;
if(sa[i-1]>len[j-1]&&sa[i-1]<len[j]) cnt+=(vis[j]?0:1),vis[j]=true;
}
}
else
{
if(cnt>k/2) ans[++size] = sa[i-1];
cnt = 0;
MEM(vis,false);
}
}
if(cnt>k/2) ans[++size] = sa[n];
if(size)
{
ans[0] = size;
return 1;
}
return 0;
}
int main()
{
int n,k,i,j,flag = 0;
while(~scanf("%d",&k),k)
{
n = 0;
size = 0;
for(i = 1; i<=k; i++)
{
scanf("%s",str+n);
for(; str[n]!=‘\0‘; n++)
s[n] = str[n];
s[n] = ‘#‘+i;
len[++size] = n;
n++;
}
s[n-1] = 0;
getsa(s,sa,n,255);
getheight(s,n-1);
int l=1,r=n,mid;
while(l<=r)
{
mid = (l+r)/2;
if(check(mid,n,k)) l = mid+1;
else r = mid-1;
}
if(flag)
puts("");
flag = 1;
if(l==1)
puts("?
");
else
{
for(i = 1; i<=ans[0]; i++)
{
for(j = ans[i]; j<ans[i]+l-1; j++)
printf("%c",s[j]);
puts("");
}
}
}
return 0;
}