题目大意：给出n个字符串，让你求出最长的子串，如果有多个按照字典序顺序输出。

解题思路：将n个字符串连起来，中间需要隔开，然后我们二分枚举字符串的长度，求最长的长度，如果多个需要按照字典序保存起来，最后输出答案就可以了。时间复杂度是：O(n*log(n))。

Description

Input

Output

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;
}

POJ 3294 Life Forms(后缀数组求k个串的最长子串)