POJ3294--Life Forms

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?

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <vector>using namespace std;#define maxn 140080#define inf 0x3f3f3f3fint key[maxn],belong[140080];char str[maxn];bool vis[120];int sa[maxn],t[maxn],t2[maxn],c[maxn];int height[maxn],Rank[maxn];/*用SA模板注意在最后添加一个比所有字符都小的字符。key[n] = 0;build_sa(key,n+1,m);getHeight(key,n+1);显然sa[0] 就是最后那个位置。。。height[i] 表示 sa[i] 和 sa[i-1] 的最长公共前缀。。*/void build_sa(int * s,int n,int m){int i,*x = t,*y = t2;for(i = 0;i < m;i++)c[i] = 0;for(i = 0;i < n;i++)c[ x[i] = s[i] ]++;for(i = 1;i < m;i++)c[i] += c[i-1];for(i = n-1;i >= 0;i--)sa[--c[x[i]]] = i;for(int k = 1;k <= n;k <<= 1){int p = 0;for(i = n - k;i < n;i++)y[p++] = i;for(i = 0;i < n;i++)if(sa[i] >= k)y[p++] = sa[i] - k;for(i = 0;i < m;i++)c[i] = 0;for(i = 0;i < n;i++)c[ x[y[i]] ]++;for(i = 0;i < m;i++)c[i] += c[i-1];for(i = n-1;i >= 0;i--)sa[--c[x[y[i]]]] = y[i];//根据sa和y数组计算新的数y组swap(x,y);p = 1;x[sa[0]] = 0;for(i = 1;i < n;i++)x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1] + k] == y[sa[i] + k] ? p-1:p++;if(p >= n)break;m = p;}}void getHeight(int * s,int n){int i,j,k = 0;for(i = 0;i < n;i++)Rank[sa[i]] = i;for(i = 0;i < n;i++){if(k) k--;int j = sa[Rank[i]-1];while(s[i+k] == s[j+k])k++;height[Rank[i]] = k;}}inline int max(int a,int b){return a>b?a:b;}bool Judge(int n,int len,int num){int cnt = 0;memset(vis,0,sizeof(vis));vis[0] = 1;if(!vis[belong[sa[0]]])cnt++;vis[belong[sa[0]]] = 1;for(int i = 1;i < n;i++){if(height[i] >= len){if(!vis[belong[sa[i]]])cnt++;}else{memset(vis,0,sizeof(vis));cnt = 0;vis[0] = 1;if(!vis[belong[sa[i]]])cnt++;}vis[belong[sa[i]]] = 1;if(cnt >= num)return 1;}return 0;}//////Judge 没有问题。。。void Print(int n,int len,int num){int cnt = 0;memset(vis,0,sizeof(vis));vis[0] = 1;if(!vis[belong[sa[0]]])cnt++;vis[belong[sa[0]]] = 1;for(int i = 1;i < n;i++){if(height[i] < len){if(cnt >= num){for(int j = sa[i-1];j < sa[i-1] + len;j++)printf("%c",key[j] - 15);printf("\n");}cnt = 0;memset(vis,0,sizeof(vis));vis[0] = 1;}if(!vis[belong[sa[i]]])cnt++;vis[belong[sa[i]]] = 1;}if(cnt >= num){for(int i = sa[n-1];i < sa[n-1] + len;i++)printf("%c",key[i] - 15);printf("\n");}}int main(){//freopen("in.txt","r",stdin);int n;int f = 0;while(scanf("%d",&n) != EOF && n){if(f)printf("\n");else f = 1;memset(belong,0,sizeof(belong));int pos = 0,scat = 1;int l = 0,r = 0;for(int i = 0;i < n;i++){scanf("%s",str);int len = strlen(str);r = max(len,r);for(int j = pos;j < pos + len;j++){key[j] = (int)str[j - pos] + 15;belong[j] = i + 1;}key[pos + len] = scat++;pos += len + 1;}if(n == 1){printf("%s\n",str);continue;}key[pos - 1] = 0;build_sa(key,pos,150);getHeight(key,pos);int maxlen = 0;while(l <= r){int mid = (l + r) >> 1;if(Judge(pos,mid,n/2+1)){maxlen = mid;l = mid + 1;}else r = mid - 1;}if(maxlen == 0){printf("?\n");}else Print(pos,maxlen,n/2+1);}return 0;}