題目大意：給出n個字符串，讓你求出最長的子串，如果有多個按照字典序順序輸出。
解題思路：將n個字符串連起來，中間需要隔開，然後我們二分枚舉字符串的長度，求最長的長度，如果多個需要按照字典序保存起來，最後輸出答案就可以了。時間複雜度是：O(n*log(n))。
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 10275 | Accepted: 2822 |
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 ?
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-9
///#define M 1000100
///#define LL __int64
#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)
#define mod 1000000007
#define Read() freopen("autocomplete.in","r",stdin)
#define Write() freopen("autocomplete.out","w",stdout)
#define Cin() ios::sync_with_stdio(false)
using namespace std;
inline int read()
{
char ch;
bool flag = false;
int a = 0;
while(!((((ch = getchar()) >= '0') && (ch <= '9')) || (ch == '-')));
if(ch != '-')
{
a *= 10;
a += ch - '0';
}
else
{
flag = true;
}
while(((ch = getchar()) >= '0') && (ch <= '9'))
{
a *= 10;
a += ch - '0';
}
if(flag)
{
a = -a;
}
return a;
}
void write(int a)
{
if(a < 0)
{
putchar('-');
a = -a;
}
if(a >= 10)
{
write(a / 10);
}
putchar(a % 10 + '0');
}
const int maxn = 200010;
int wa[maxn], wb[maxn], wv[maxn], ws1[maxn];
int sa[maxn];
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 *sa, int n, int m)
{
int i, j, p, *x = wa, *y = wb;
for(i = 0; i < m; i++) ws1[i] = 0;
for(i = 0; i < n; i++) ws1[x[i] = r[i]]++;
for(i = 1; i < m; i++) ws1[i] += ws1[i-1];
for(i = n-1; i >= 0; i--) sa[--ws1[x[i]]] = i;
for(j = 1, p = 1; p < n; j <<= 1, 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++) ws1[i] = 0;
for(i = 0; i < n; i++) ws1[wv[i]]++;
for(i = 1; i < m; i++) ws1[i] += ws1[i-1];
for(i = n-1; i >= 0; i--) sa[--ws1[wv[i]]] = y[i];
for(swap(x, y), p = 1, x[sa[0]] = 0, i = 1; i < n; i++)
x[sa[i]] = cmp(y, sa[i-1], sa[i], j)?p-1:p++;
}
return ;
}
int rank[maxn], height[maxn];
void calheight(int *r, int *sa, int n)
{
int i, j, k = 0;
for(i = 1; i <= n; i++) rank[sa[i]] = i;
for(i = 0; i < n; height[rank[i++]] = k)
for(k?k--:0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k++);
return;
}
char str1[maxn], str2[maxn];
int seq[maxn];
int hash[maxn];
char str[110][1010];
struct node
{
int pos;
int s;
int len;
}f[maxn], xf[maxn];
int xans;
int vis[110];
int Find(int x)
{
int s = hash[x];
for(int i = 0; i < s; i++)
{
int len = strlen(str[i]);
x -= len;
}
x -= s;
return x;
}
bool judge(int mid, int n, int m)
{
int ans = 0;
int sx;
for(int i = 2; i <= n; i++)
{
memset(vis, 0, sizeof(vis));
sx = 1;
vis[hash[sa[i-1]]] = 1;
while(height[i] >= mid)
{
if(!vis[hash[sa[i]]])
{
vis[hash[sa[i]]] = 1;
++sx;
}
i++;
}
if(sx*2 > m)
{
xf[ans].len = mid;
xf[ans].pos = Find(sa[i-1]);
xf[ans++].s = hash[sa[i-1]];
}
}
if(ans)
{
xans = 0;
for(int i = 0; i < ans; i++)
f[xans++] = xf[i];
}
return ans;
}
void Del(int n, int len, int m)
{
int l = 1;
int r = len;
xans = 0;
while(l <= r)
{
int mid = (l+r)>>1;
if(judge(mid, n, m)) l = mid+1;
else r = mid-1;
}
if(!xans)
{
cout<<"?"<<endl;
return;
}
for(int i = 0; i < xans; i++)
{
for(int j = f[i].pos, k = 0; k < f[i].len; k++, j++)
cout<<str[f[i].s][j];
puts("");
}
}
int main()
{
int n;
int flag = 0;
while(~scanf("%d", &n) && n)
{
memset(hash, -1, sizeof(hash));
int ans = 0;
int Min = maxn;
for(int i = 0; i < n; i++)
{
scanf("%s",str[i]);
int len = strlen(str[i]);
Min = min(Min, len);
for(int j = 0; j < len; j++)
{
seq[ans] = str[i][j];
hash[ans++] = i;
}
seq[ans++] = 200+i;
}
seq[ans] = 0;
da(seq, sa, ans+1, 310);
calheight(seq, sa, ans);
if(!flag) flag = 1;
else puts("");
Del(ans, Min, n);
}
return 0;
}