POJ1226---Substrings（后缀数组+二分）

Description
You are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings.

Input
The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains a single integer n (1 <= n <= 100), the number of given strings, followed by n lines, each representing one string of minimum length 1 and maximum length 100. There is no extra white space before and after a string.

Output
There should be one line per test case containing the length of the largest string found.

Sample Input

2
3
ABCD
BCDFF
BRCD
2
rose
orchid

Sample Output

2
2

Source
Tehran 2002 Preliminary

把串，反串都连在一起，求后缀数组，然后二分，给后缀数组分组就行了

/*************************************************************************    > File Name: POJ1226.cpp    > Author: ALex    > Mail: [e-mail]     > Created Time: 2015年04月07日 星期二 16时45分07秒 ************************************************************************/#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <queue>#include <stack>#include <map>#include <bitset>#include <set>#include <vector>using namespace std;const double pi = acos(-1.0);const int inf = 0x3f3f3f3f;const double eps = 1e-15;typedef long long LL;typedef pair <int, int> PLL;int pos[22000];class SuffixArray{    public:        static const int N = 22000;        int init[N];        int X[N];        int Y[N];        int Rank[N];        int sa[N];        int height[N];        int buc[N];        int LOG[N];        int dp[N][20];        int size;        bool vis[110];        void clear()        {            size = 0;        }        void insert(int n)        {            init[size++] = n;        }        bool cmp(int *r, int a, int b, int l)        {            return (r[a] == r[b] && r[a + l] == r[b + l]);        }        void getsa(int m = 256) //m一般为最大值+1        {            init[size] = 0;            int l, p, *x = X, *y = Y, n = size + 1;            for (int i = 0; i < m; ++i)            {                buc[i] = 0;            }            for (int i = 0; i < n; ++i)            {                ++buc[x[i] = init[i]];            }            for (int i = 1; i < m; ++i)            {                buc[i] += buc[i - 1];            }            for (int i = n - 1; i >= 0; --i)            {                sa[--buc[x[i]]] = i;            }            for (l = 1, p = 1; l <= n && p < n; m = p, l *= 2)            {                p = 0;                for (int i = n - l; i < n; ++i)                {                    y[p++] = i;                }                for (int i = 0; i < n; ++i)                {                    if (sa[i] >= l)                    {                        y[p++] = sa[i] - l;                    }                }                for (int i = 0; i < m; ++i)                {                    buc[i] = 0;                }                for (int i = 0; i < n; ++i)                {                    ++buc[x[y[i]]];                }                for (int i = 1; i < m; ++i)                {                    buc[i] += buc[i - 1];                }                for (int i = n - 1; i >= 0; --i)                {                    sa[--buc[x[y[i]]]] = y[i];                }                int i;                for (swap(x, y), x[sa[0]] = 0, p = 1, i = 1; i < n; ++i)                {                     x[sa[i]] = cmp(y, sa[i - 1], sa[i], l) ? p - 1 : p++;                 }            }        }        void getheight()        {            int h = 0, n = size;            for (int i = 0; i <= n; ++i)            {                Rank[sa[i]] = i;            }            height[0] = 0;            for (int i = 0; i < n; ++i)            {                if (h > 0)                {                    --h;                }                int j =sa[Rank[i] - 1];                for (; i + h < n && j + h < n && init[i + h] == init[j + h]; ++h);                height[Rank[i] - 1] = h;            }        }           //预处理每一个数字的对数，用于rmq，常数优化        void initLOG()        {            LOG[0] = -1;            for (int i = 1; i < N; ++i)            {                LOG[i] = (i & (i - 1)) ? LOG[i - 1] : LOG[i - 1] + 1;            }        }        void initRMQ()        {            initLOG();            int n = size;            int limit;            for (int i = 0; i < n; ++i)            {                dp[i][0] = height[i];            }            for (int j = 1; j <= LOG[n]; ++j)            {                limit = (n - (1 << j));                for (int i = 0; i <= limit; ++i)                {                    dp[i][j] = min(dp[i][j - 1], dp[i + (1 << j) - 1][j - 1]);                }            }        }        int LCP(int a, int b)        {            int t;            a = Rank[a];            b = Rank[b];            if (a > b)            {                swap(a, b);            }--b;            t = LOG[b - a + 1];            return min(dp[a][t], dp[b - (1 << t) + 1][t]);        }        bool check(int k, int n)        {            int cnt = 1;            memset(vis, 0, sizeof(vis));            vis[pos[sa[1]]] = 1;            for (int i = 1; i < size; ++i)            {                if (height[i] >= k)                {                    if (!vis[pos[sa[i + 1]]])                    {                        ++cnt;                        vis[pos[sa[i + 1]]] = 1;                    }                }                else                {                    if (cnt == n)                    {                        return 1;                    }                    memset(vis, 0, sizeof(vis));                    vis[pos[sa[i + 1]]] = 1;                    cnt = 1;                }            }            return 0;        }        void solve(int n)        {            int l = 1, r = size, mid, ans = 0;            while (l <= r)            {                mid = (l + r) >> 1;                if (check(mid, n))                {                    l = mid + 1;                    ans = mid;                }                else                {                    r = mid - 1;                }            }            printf("%d\n", ans);        }}SA;char str[120];int main(){    int t;    scanf("%d", &t);    while (t--)    {        int n, maxs = 256, cnt = 0;        SA.clear();        scanf("%d", &n);        if (n == 1)        {            scanf("%s", str);            int len = strlen(str);            printf("%d\n", len);            continue;        }        for (int i = 1; i <= n; ++i)        {            scanf("%s", str);            int len = strlen(str);            for (int j = 0; j < len; ++j)            {                SA.insert((int)str[j]);                pos[cnt++] = i;            }            SA.insert(maxs++);            pos[cnt++] = 0;            reverse(str, str + len);            for (int j = 0; j < len; ++j)            {                SA.insert((int)str[j]);                pos[cnt++] = i;            }            pos[cnt++] = 0;            SA.insert(maxs++);        }        SA.getsa(maxs);        SA.getheight();        SA.solve(n);    }    return 0;}