Life Forms

算是挺简单的。有一些坑点。
我们通过分隔符将字符串们连接起来。
注意分隔符不可以相等，否则在匹配时分隔符可能被算上。
然后我们二分答案长度，进行height分组，判断是否可行。
得到正确的长度后我们再去找子串，只要找到分组就好了，然后随便记录下分组中的一个坐标。
之后将这些坐标按照ranks[i]序排序就好了。
之后输出，每一个输出len长
在二分判断时，我们要注意判断时n的奇偶。最方便的是乘2看是否大于n

代码如下：

#include<iostream>
#include<algorithm>
#include<set>
#include<vector>
using namespace std;
const int max_n = 1e5 + 200;
int a[max_n];
int ranks[max_n], SA[max_n], height[max_n];
int wa[max_n], wb[max_n], wvarr[max_n], wsarr[max_n];
inline int cmp(int* r, int a, int b, int l) {
    return r[a] == r[b] && r[a + l] == r[b + l];
}
inline void get_sa(int* r, int* sa, int n, int m) {
    //r:原数组
    //sa:SA
    //n:原数组长度
    //m:原数组种类数，用于基数排序
    ++n;
    int i, j, p, * x = wa, * y = wb, * t;
    for (i = 0; i < m; ++i) wsarr[i] = 0;
    for (i = 0; i < n; ++i) wsarr[x[i] = r[i]]++;
    for (i = 1; i < m; ++i) wsarr[i] += wsarr[i - 1];
    for (i = n - 1; i >= 0; --i) sa[--wsarr[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) wvarr[i] = x[y[i]];
        for (i = 0; i < m; ++i) wsarr[i] = 0;
        for (i = 0; i < n; ++i) wsarr[wvarr[i]]++;
        for (i = 1; i < m; ++i) wsarr[i] += wsarr[i - 1];
        for (i = n - 1; i >= 0; --i) sa[--wsarr[wvarr[i]]] = y[i];
        for (t = x, x = y, y = t, 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++;
    }
}//求解高度数组，height[i]指排名
void get_height(int* r, int* sa, int n) {
    int i, j, k = 0;
    for (i = 0; i <= n; ++i) ranks[sa[i]] = i;
    for (i = 0; i < n; height[ranks[i++]] = k)
        for (k ? k-- : 0, j = sa[ranks[i] - 1]; r[i + k] == r[j + k]; k++);
    return;
}
int n;
int col[max_n];
bool check(int mid,int tot) {
    set<int>myset;
    for (int i = 2;i <= tot;++i) {
        if (myset.size() * 2 > n)return true;
        if (height[i] < mid)myset.clear();
        else {
            myset.insert(col[SA[i]]);
            myset.insert(col[SA[i - 1]]);
        }
    }return false;
}
bool ccmp(int aa, int bb) { return ranks[aa] < ranks[bb]; }
int main(){
    ios::sync_with_stdio(0);
    while (cin >> n) {
        if (n == 0)break;
        if (n == 1) {
            string s;cin >> s;
            cout << s << "\n\n";
            continue;
        }
        fill(col, col + max_n, -1);
        int tot = 0;int co = 0;
        for (int i = 1;i <= n;++i) {
            string s;cin >> s;++co;
            for (int j = 0;j < s.size();++j) {
                a[tot++] = s[j] - 'a' + 1;
                col[tot - 1] = co;
            }a[tot++] = 30 + i;
        }--tot;a[tot] = 0;
        get_sa(a, SA, tot, 200);
        get_height(a, SA, tot);
        int lft = 1;int rght = 1000;
        while (lft <= rght) {
            int mid = (lft + rght) >> 1;
            if (check(mid, tot))lft = mid + 1;
            else rght = mid - 1;
        }
        int len = lft - 1;
        if (len == 0) {
            cout << "?\n\n";
            continue;
        }vector<int> ans;
        set<int> myset;int id = -1;
        for (int i = 2;i <= tot;++i) {
            if (height[i] >= len) {
                id = SA[i];
                myset.insert(col[SA[i]]);
                myset.insert(col[SA[i - 1]]);
            }
            else {
                if (myset.size() * 2 > n)
                    ans.push_back(id);
                myset.clear();
            }
        }if (myset.size() * 2 > n)
            ans.push_back(id);
        sort(ans.begin(), ans.end(), ccmp);
        for (int i = 0;i < ans.size();++i) {
            int id = ans[i];
            for (int j = 0;j < len;++j)
                cout << (char)(a[id + j] + 'a' - 1);
            cout << endl;
        }cout << endl;
    }
}