[hdu4436 str2int]后缀自动机SAM（或后缀数组SA）

题意：给n个数字串，求它们的所有不包含前导0的不同子串的值之和

思路：把数字串拼接在一起，构造SAM，然后以每个状态的长度len作为特征值从小到大排序，从前往后处理每个状态，相当于按拓扑序在图上合并计算答案。

#include <bits/stdc++.h>
using namespace std;
#define X first
#define Y second
#define pb(x) push_back(x)
#define mp(x, y) make_pair(x, y)
#define all(a) (a).begin(), (a).end()
#define mset(a, x) memset(a, x, sizeof(a))
#define mcpy(a, b) memcpy(a, b, sizeof(b))
#define cas() int T, cas = 0; cin >> T; while (T --)
template<typename T>bool umax(T&a, const T&b){return a<b?(a=b,true):false;}
template<typename T>bool umin(T&a, const T&b){return b<a?(a=b,true):false;}
typedef long long ll;
typedef pair<int, int> pii;

#ifndef ONLINE_JUDGE
    #include "local.h"
#endif

const int mod = 2012;

const int N = 2e5;

class SAM {
public:
    void init() {
        memset(node, 0, sizeof(node));
        sz = last = 0;
        node[0].len = 0;
        node[0].link = -1;
        sz ++;
    }
    void add(char c) {
        int cur = sz ++;
        node[cur].len = node[last].len + 1;
        int p;
        for (p = last; ~p && !node[p].next[c]; p = node[p].link) {
            node[p].next[c] = cur;
        }
        if (p == -1) node[cur].link = 0;
        else {
            int q = node[p].next[c];
            if (node[p].len + 1 == node[q].len) node[cur].link = q;
            else {
                int clone = sz ++;
                node[clone] = node[q];
                node[clone].len = node[p].len + 1;
                for (; ~p && node[p].next[c] == q; p = node[p].link) {
                    node[p].next[c] = clone;
                }
                node[q].link = node[cur].link = clone;
            }
        }
        last = cur;
    }
    int c[N << 1], p[N << 1];
    int getAns() {
        mset(c, 0);
        for (int i = 0; i < sz; i ++) c[node[i].len] ++;
        for (int i = 1; i <= sz; i ++) c[i] += c[i - 1];
        for (int i = 0; i < sz; i ++) p[-- c[node[i].len]] = i;
        int ans = 0;
        node[0].cnt = 1;
        for (int i = 0; i < sz; i ++) {
            int cur = p[i];
            for (int j = 0; j < 10; j ++) {
                if (!cur && !j || !node[cur].next[j]) continue;
                int next = node[cur].next[j];
                node[next].sum = (node[next].sum + node[cur].sum * 10 + j * node[cur].cnt) % mod;
                node[next].cnt = (node[next].cnt + node[cur].cnt) % mod;
            }
            ans = (ans + node[cur].sum) % mod;
        }
        return ans;
    }
private:
    const static int SZ = 11;
    struct State {
        int len, link;
        int next[SZ];
        int sum, cnt;
    };
    State node[N << 1];
    int sz, last;
};
SAM sam;
char s[123456];

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
    int n;
    while (cin >> n) {
        sam.init();
        for (int i = 0; i < n; i ++) {
            scanf("%s", s);
            for (int j = 0; s[j]; j ++) {
                sam.add(s[j] - '0');
            }
            sam.add(10);
        }
        cout << sam.getAns() << endl;
    }
    return 0;
}