【SPOJ】LCS2（SAM）

Description

求多个串的LCS

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Solution

建出一个串的SAM，然后将剩下的每一个串在上面跑一遍，每个节点取对于每个串匹配出来的最小值，答案就是所有节点的最大值。
记得要用子节点更新父节点

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Code

/************************************************
 * Au: Hany01
 * Date: May 4th, 2018
 * Prob: SPOJ LCS2
 * Email: [email protected]
************************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read()
{
    register int _, __; register char c_;
    for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
    for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 100005;

struct SAM
{
    int len, fa, ch[26];
}t[maxn << 1];
int tot = 1, las = 1;
int n, Min[maxn << 1], tmp[maxn << 1], per[maxn << 1], c[maxn];
char s[maxn];

inline void extend(int c)
{
    int np = ++ tot, p = las;
    las = tot, t[np].len = t[p].len + 1;
    while (p && !t[p].ch[c]) t[p].ch[c] = np, p = t[p].fa;
    if (!p) t[np].fa = 1;
    else {
        register int q = t[p].ch[c];
        if (t[q].len == t[p].len + 1) t[np].fa = q;
        else {
            register int nq = ++ tot;
            t[nq] = t[q], t[nq].len = t[p].len + 1;
            t[q].fa = t[np].fa = nq;
            while (p && t[p].ch[c] == q) t[p].ch[c] = nq, p = t[p].fa;
        }
    }
}

int main()
{
#ifdef hany01
    File("LCS2");
#endif

    scanf("%s", s);
    rep(i, n = strlen(s)) extend(s[i] - 97);

    For(i, 1, tot) Min[i] = t[i].len;
    For(i, 1, tot) ++ c[t[i].len];
    For(i, 1, n) c[i] += c[i - 1];
    For(i, 1, tot) per[c[t[i].len] --] = t[i].len;

    while (scanf("%s", s) != EOF)
    {
        int u = 1, now = 0;
        Set(tmp, 0);
        rep(i, strlen(s)) {
            s[i] -= 97;
            if (t[u].ch[s[i]]) ++ now, u = t[u].ch[s[i]];
            else {
                while (u && !t[u].ch[s[i]]) u = t[u].fa;
                if (!u) now = 0, u = 1;
                else now = t[u].len + 1, u = t[u].ch[s[i]];
            }
            chkmax(tmp[u], now);
        }
        Fordown(i, tot, 1) chkmax(tmp[t[per[i]].fa], tmp[per[i]]);
        For(i, 1, tot) chkmin(Min[i], tmp[i]);
    }
    static int Max = 0;
    For(i, 2, tot) chkmax(Max, Min[i]);
    printf("%d\n", Max);

    return 0;
}
//东风夜放花千树。更吹落、星如雨。
//    -- 辛弃疾《青玉案·元夕》