SPOJ：LCS2

题面

给定一些字符串，求出它们的最长公共子串 输入格式 输入至多 (10) 行，每行包含不超过 (100000)个的小写字母，表示一个字符串 输出格式 一个数，最长公共子串的长度 若不存在最长公共子串，请输出 (0)

Sol

一个串建立(sam)
每个串在上面匹配
每个点匹配的长度可以由后继转移过来
拓扑序上(DP)

# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;

template <class Int>
IL void Input(RG Int &x){
    RG int z = 1; RG char c = getchar(); x = 0;
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    x *= z;
}

const int maxn(2e5 + 5);

int n, trans[26][maxn], fa[maxn], len[maxn], tot = 1, last = 1, ans, f[maxn], g[maxn];
int id[maxn], t[maxn];
char s[maxn];

IL void Extend(RG int c){
    RG int p = last, np = ++tot; last = np;
    len[np] = len[p] + 1;
    while(p && !trans[c][p]) trans[c][p] = np, p = fa[p];
    if(!p) fa[np] = 1;
    else{
        RG int q = trans[c][p];
        if(len[q] == len[p] + 1) fa[np] = q;
        else{
            RG int nq = ++tot;
            fa[nq] = fa[q], len[nq] = len[p] + 1;
            for(RG int i = 0; i < 26; ++i) trans[i][nq] = trans[i][q];
            fa[q] = fa[np] = nq;
            while(p && trans[c][p] == q) trans[c][p] = nq, p = fa[p];
        }
    }
}

int main(RG int argc, RG char* argv[]){
    scanf(" %s", s), n = strlen(s);
    for(RG int i = 0; i < n; ++i) Extend(s[i] - 'a');
    for(RG int i = 1; i <= tot; ++i) ++t[g[i] = len[i]];
    for(RG int i = 1; i <= tot; ++i) t[i] += t[i - 1];
    for(RG int i = 1; i <= tot; ++i) id[t[len[i]]--] = i;
    while(scanf(" %s", s) != EOF){
        n = strlen(s);
        for(RG int i = 1; i <= tot; ++i) f[i] = 0;
        for(RG int i = 0, nw = 1, cnt = 0; i < n; ++i){
            RG int c = s[i] - 'a';
            if(trans[c][nw]) ++cnt, nw = trans[c][nw];
            else{
                while(nw && !trans[c][nw]) nw = fa[nw], cnt = len[nw];
                if(!nw) nw = 1, cnt = 0;
                else cnt++, nw = trans[c][nw];
            }
            f[nw] = max(f[nw], cnt);
        }
        for(RG int i = tot; i; --i) f[fa[id[i]]] = max(f[fa[id[i]]], f[id[i]]);
        for(RG int i = 1; i <= tot; ++i) g[i] = min(g[i], f[i]);
    }
    for(RG int i = 1; i <= tot; ++i) ans = max(ans, g[i]);
    printf("%d
", ans);
    return 0;
}