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: [email protected] > 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;
}