| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 12484 | Accepted: 3502 |
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.
/*
POJ 3294 n个串中至少一半的串共享的最长公共子串
求的是最长公共子串，所以考虑 二分答案len+判断
因为要判断是否为x个串共享所以对height进行分组,即height数组中各个连续≥len
的集合，然后对每个组进行判断,看书否能找到x+1个不同的来源。
满足条件就记录 子串的起始位置和长度
1.串之间的间隔符号不能相同
2.因为有100个串,所以已经占据了0-99,所以字符串的信息转换成int的时候
必需是从100开始
hhh-2016-03-17 19:04:50
*/
#include <algorithm>
#include <cmath>
#include <queue>
#include <iostream>
#include <cstring>
#include <map>
#include <cstdio>
#include <vector>
#include <functional>
#define lson (i<<1)
#define rson ((i<<1)|1)
using namespace std;
typedef long long ll;
const int maxn = 101000;
int t1[maxn],t2[maxn],c[maxn];
bool cmp(int *r,int a,int b,int l)
{
return r[a]==r[b] &&r[l+a] == r[l+b];
}
void get_sa(int str[],int sa[],int Rank[],int height[],int n,int m)
{
n++;
int p,*x=t1,*y=t2;
for(int i = 0; i < m; i++) c[i] = 0;
for(int i = 0; i < n; i++) c[x[i] = str[i]]++;
for(int i = 1; i < m; i++) c[i] += c[i-1];
for(int i = n-1; i>=0; i--) sa[--c[x[i]]] = i;
for(int j = 1; j <= n; j <<= 1)
{
p = 0;
for(int i = n-j; i < n; i++) y[p++] = i;
for(int i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i]-j;
for(int i = 0; i < m; i++) c[i] = 0;
for(int i = 0; i < n; i++) c[x[y[i]]]++ ;
for(int i = 1; i < m; i++) c[i] += c[i-1];
for(int i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
swap(x,y);
p = 1;
x[sa[0]] = 0;
for(int i = 1; i < n; i++)
x[sa[i]] = cmp(y,sa[i-1],sa[i],j)? p-1:p++;
if(p >= n) break;
m = p;
}
int k = 0;
n--;
for(int i = 0; i <= n; i++)
Rank[sa[i]] = i;
for(int i = 0; i < n; i++)
{
if(k) k--;
int j = sa[Rank[i]-1];
while(str[i+k] == str[j+k]) k++;
height[Rank[i]] = k;
}
}
int Rank[maxn];
int sa[maxn];
int str[maxn],height[maxn];
char s[1010];
char allstr[maxn];
int anslen,anspos[maxn];
int ansnum,vis[110];
int id[maxn];
bool judge(int len,int k,int n,int l,int r)
{
int num = 0;
memset(vis,0,sizeof(vis));
for(int i = l; i <= r; i++)
{
if(height[i] >= len)
{
if(!vis[id[sa[i-1]]])
{
vis[id[sa[i-1]]] = 1;
num ++;
}
if(!vis[id[sa[i]]])
{
vis[id[sa[i]]] = 1;
num ++;
}
if(num > k)
return 1;
}
}
return 0;
}
bool can(int len,int k,int n)
{
int l=2,r=2;
int flag = 0;
ansnum = 0;
for(int i = 2; i <= n; i++)
{
if(height[i]>=len)
r++;
else
{
if(judge(len,k,n,l,r))
{
anspos[ansnum++] = sa[l];
flag =1;
}
l = i,r = i;
}
}
if(judge(len,k,n,l,r) && l < r)
{
anspos[ansnum++] = sa[l];
flag =1;
}
return flag;
}
int main()
{
int k,n;
while(scanf("%d",&n) != EOF && n)
{
int len=0;
int tot = 0;
for(int i = 0; i< n; i++)
{
scanf("%s",s);
for(int j = 0; s[j]!=' '; j++)
{
id[tot] = i;
allstr[tot] = s[j];
str[tot++] = s[j]-'a'+100;
}
len=max(len,(int)strlen(s));
id[tot] = i,allstr[tot]='$';
str[tot++]=i;
}
str[tot] = 0;
get_sa(str,sa,Rank,height,tot,128);
int k = n/2;
int ans = 0;
int l=1,r=len;
while(l <= r)
{
int mid =(l+r)>>1;
if(can(mid,k,tot))
{
l = mid+1;
anslen = mid;
ans = ansnum;
}
else
r = mid-1;
}
if(!ans)
printf("?
");
else
{
//cout<<ans<<endl;
for(int i = 0; i < ans; i++)
{
for(int j = 0; j<anslen; j++)
printf("%c",allstr[anspos[i]+j]);
printf("
");
}
}
printf("
");
}
return 0;
}