【bzoj3277&&3474】串

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*题目描述：
字符串是oi界常考的问题。现在给定你n个字符串，询问每个字符串有多少子串（不包括空串）是所有n个字符串中至少k个字符串的子串（注意包括本身）。

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*输入：
第一行两个整数n，k。接下来n行每行一个字符串。

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*输出：
输出一行n个整数，第i个整数表示第i个字符串的答案。

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*样例输入：
3 1
abc
a
ab

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*样例输出：
6 1 3

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*提示：
对于100%的数据，n,k,l<=100000

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*来源：
后缀数组

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*题解：
广义后缀自动机。建完广义后缀自动机后，统计一下某个节点在所有字符串中出现的次数，对于次数大于等于k的节点统计一下答案。

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*代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

#ifdef WIN32
    #define LL "%I64d"
#else
    #define LL "%lld"
#endif

#ifdef CT
    #define debug(...) printf(__VA_ARGS__)
    #define setfile() 
#else
    #define debug(...)
    #define filename ""
    #define setfile() freopen(filename".in", "r", stdin); freopen(filename".out", "w", stdout);
#endif

#define R register
#define getc() (S == T && (T = (S = B) + fread(B, 1, 1 << 15, stdin), S == T) ? EOF : *S++)
#define dmax(_a, _b) ((_a) > (_b) ? (_a) : (_b))
#define dmin(_a, _b) ((_a) < (_b) ? (_a) : (_b))
#define cmax(_a, _b) (_a < (_b) ? _a = (_b) : 0)
#define cmin(_a, _b) (_a > (_b) ? _a = (_b) : 0)
char B[1 << 15], *S = B, *T = B;
inline int FastIn()
{
    R char ch; R int cnt = 0; R bool minus = 0;
    while (ch = getc(), (ch < '0' || ch > '9') && ch != '-') ;
    ch == '-' ? minus = 1 : cnt = ch - '0';
    while (ch = getc(), ch >= '0' && ch <= '9') cnt = cnt * 10 + ch - '0';
    return minus ? -cnt : cnt;
}
#define maxn 100010
struct sam
{
    sam *next[26], *fa;
    int val, last_vis, c;
    bool vis;
    long long sum;
}mem[maxn << 1], *tot = mem;
inline sam *extend(R sam *p, R int c)
{
    if (p -> next[c])
    {
        R sam *q = p -> next[c];
        if (q -> val == p -> val + 1)
            return q;
        else
        {
            R sam *nq = ++tot;
            memcpy(nq -> next, q -> next, sizeof nq -> next);
            nq -> val = p -> val + 1;
            nq -> fa = q -> fa;
            q -> fa = nq;
            for ( ; p && p -> next[c] == q; p = p -> fa)
                p -> next[c] = nq;
            return nq;
        }
    }
    R sam *np = ++tot;
    np -> val = p -> val + 1;
    for ( ; p && !p -> next[c]; p = p -> fa) p -> next[c] = np;
    if (!p)
        np -> fa = mem;
    else
    {
        R sam *q = p -> next[c];
        if (q -> val == p -> val + 1)
            np -> fa = q;
        else
        {
            R sam *nq = ++tot;
            memcpy(nq -> next, q -> next, sizeof nq -> next);
            nq -> val = p -> val + 1;
            nq -> fa = q -> fa;
            q -> fa = np -> fa = nq;
            for ( ; p && p -> next[c] == q; p = p -> fa)
                p -> next[c] = nq;
        }
    }
    return np;
}
void get_ans(R sam *x)
{
    if (x == mem || x -> vis) return;
    x -> vis = 1; get_ans(x -> fa); x -> sum += x -> fa -> sum;
}
char str[maxn], tot_str[maxn];
int left[maxn], right[maxn];
int main()
{
//  setfile();
    R int n, k;
    scanf("%d%d", &n, &k);
    R int tot_len = 0;
    for (R int i = 1; i <= n; ++i)
    {
        scanf("%s", str);
        R sam* x = mem;
        R int len = strlen(str);
        left[i] = tot_len;
        right[i] = tot_len = len + tot_len - 1; ++tot_len;
        memcpy(tot_str + left[i], str, len * sizeof(char));
        for (R int j = 0; j < len; ++j)
            x = extend(x, str[j] - 'a');
    }
    for (R int i = 1; i <= n; ++i)
    {
        R sam *x = mem, *t;
        for (R int j = left[i]; j <= right[i]; ++j)
        {
            x = x -> next[tot_str[j] - 'a'];
            for (t = x; t && t -> last_vis != i; t = t -> fa)
                t -> last_vis = i, t -> c++;
        }
    }
    for (R sam *iter = mem + 1; iter <= tot; ++iter)
        iter -> sum = iter -> c >= k ? iter -> val - iter -> fa -> val : 0;
    for (R sam *iter = mem + 1; iter <= tot; ++iter)
        get_ans(iter);
    for (R int i = 1; i <= n; ++i)
    {
        R sam *x = mem; R long long ans = 0;
        for (R int j = left[i]; j <= right[i]; ++j)
            x = x -> next[tot_str[j] - 'a'], ans += x -> sum;
        printf("%lld ", ans );
    }
    return 0;
}