Blue Jeans 【POJ - 3080】【后缀数组模板+尺取区间最小值】

题目链接 求解多个串最长相同元素模板

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 给出N个串，求N个串中的出现长度最长的公共相同子串。如果有多个相同的，输出最小字典序的那个。如果长度小于3，则输出题目中所给出的。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
//#include <unordered_map>
//#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e3 + 7;
struct SA
{
    int n, m;
    int s[maxN];
    int y[maxN], x[maxN], c[maxN], sa[maxN], rk[maxN], height[maxN];
    inline void get_SA()
    {
        for(int i=1; i<=m; i++) c[i] = 0;   //桶的初始化
        for(int i=1; i<=n; i++) ++c[x[i] = s[i]];
        for(int i=2; i<=m; i++) c[i] += c[i - 1];   //利用差分前缀和的思想知道每个关键字最多是在第几名
        for(int i=n; i>=1; i--) sa[c[x[i]]--] = i;
        for(int k=1; k<=n; k<<=1)
        {
            int num = 0;
            for(int i=n - k + 1; i<=n; i++) y[++num] = i;
            for(int i=1; i<=n; i++) if(sa[i] > k) y[++num] = sa[i] - k; //是否可以作为第二关键字
            for(int i=1; i<=m; i++) c[i] = 0;
            for(int i=1; i<=n; i++) c[x[i]]++;  //因为上一次循环已经求出这次的第一关键字了
            for(int i=2; i<=m; i++) c[i] += c[i - 1];
            for(int i=n; i>=1; i--) //在同一第一关键字下，按第二关键字来排
            {
                sa[c[x[y[i]]]--] = y[i];
                y[i] = 0;
            }
            swap(x, y);
            x[sa[1]] = 1; num = 1;
            for(int i=2; i<=n; i++)
            {
                x[sa[i]] = (y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]) ? num : ++num;
            }
            if(num == n) break;
            m = num;
        }
    }
    inline void get_height()
    {
        int k = 0;
        for(int i=1; i<=n; i++) rk[sa[i]] = i;
        for(int i=1; i<=n; i++)
        {
            if(rk[i] == 1) continue;    //第一名的height为0
            if(k) k--;  //height[i] >= height[i - 1] - 1
            int j = sa[rk[i] - 1];
            while(j + k <= n && i + k <= n && s[i + k] == s[j + k]) k++;
            height[rk[i]] = k;
        }
    }
    inline void clear()
    {
        n = 0; m = 200;
    }
} sa;
struct BIT_Tree
{
    int tree[maxN << 2];
    inline void buildTree(int rt, int l, int r)
    {
        if(l == r) { tree[rt] = sa.height[l]; return; }
        int mid = HalF;
        buildTree(Lson); buildTree(Rson);
        tree[rt] = min(tree[lsn], tree[rsn]);
    }
    inline int query(int rt, int l, int r, int ql, int qr)
    {
        if(ql <= l && qr >= r) return tree[rt];
        int mid = HalF;
        if(qr <= mid) return query(QL);
        else if(ql > mid) return query(QR);
        else return min(query(QL), query(QR));
    }
} tree;
int N, len, str_end[15];;
char s[65];
int vis[15], get_person, ans_len, ans_beg, tmp_len;
int main()
{
    int T; scanf("%d", &T);
    while(T--)
    {
        sa.clear(); get_person = 0;
        scanf("%d", &N);
        for(int i=1; i<=N; i++) vis[i] = 0;
        for(int i=1; i<=N; i++)
        {
            scanf("%s", s + 1);
            len = (int)strlen(s + 1);
            for(int j=1; j<=len; j++)
            {
                sa.s[++sa.n] = s[j];
            }
            sa.s[++sa.n] = 'a' + 26 + i;
            str_end[i] = sa.n;
        }
        sa.get_SA();
        sa.get_height();
        tree.buildTree(1, 2, sa.n);
        int l = 1, r = 1, tmp_id;
        while(get_person < N)
        {
            tmp_id = (int)(lower_bound(str_end + 1, str_end + N + 1, sa.sa[r]) - str_end);
            if(!vis[tmp_id]) get_person ++;
            vis[tmp_id] ++;
            r++;
        }
        tmp_id = (int)(lower_bound(str_end + 1, str_end + N + 1, sa.sa[l]) - str_end);
        while(vis[tmp_id] > 1)
        {
            vis[tmp_id]--;
            l ++;
            tmp_id = (int)(lower_bound(str_end + 1, str_end + N + 1, sa.sa[l]) - str_end);
        }
        ans_len = tree.query(1, 2, sa.n, l + 1, r - 1);
        ans_beg = sa.sa[r - 1];
        while(r <= sa.n - N)
        {
            tmp_id = (int)(lower_bound(str_end + 1, str_end + N + 1, sa.sa[r]) - str_end);
            vis[tmp_id]++;
            tmp_id = (int)(lower_bound(str_end + 1, str_end + N + 1, sa.sa[l]) - str_end);
            while(vis[tmp_id] > 1)
            {
                vis[tmp_id]--;
                l++;
                tmp_id = (int)(lower_bound(str_end + 1, str_end + N + 1, sa.sa[l]) - str_end);
            }
            tmp_len = tree.query(1, 2, sa.n, l + 1, r);
            if(tmp_len > ans_len)
            {
                ans_len = tmp_len;
                ans_beg = sa.sa[r];
            }
            r++;
        }
        if(ans_len < 3) printf("no significant commonalities\n");
        else
        {
            for(int i=ans_beg; i<ans_beg + ans_len; i++) printf("%c", sa.s[i]); puts("");
        }
    }
    return 0;
}