HDU 4436 str2int

Problem Description

In this problem, you are given several strings that contain only digits from '0' to '9', inclusive.
An example is shown below.
101
123
The set S of strings is consists of the N strings given in the input file, and all the possible substrings of each one of them.
It's boring to manipulate strings, so you decide to convert strings in S into integers.
You can convert a string that contains only digits into a decimal integer, for example, you can convert "101" into 101, "01" into 1, et al.
If an integer occurs multiple times, you only keep one of them. 
For example, in the example shown above, all the integers are 1, 10, 101, 2, 3, 12, 23, 123.
Your task is to calculate the remainder of the sum of all the integers you get divided by 2012.

Input

There are no more than 20 test cases.
The test case starts by a line contains an positive integer N.
Next N lines each contains a string consists of one or more digits.
It's guaranteed that 1≤N≤10000 and the sum of the length of all the strings ≤100000.
The input is terminated by EOF.

Output

An integer between 0 and 2011, inclusive, for each test case.

Sample Input

5 101 123 09 000 1234567890

Sample Output

202

后缀自动机+拓扑排序

#include<cstdio>    
#include<cstring>    
#include<cmath>    
#include<algorithm>    
#include<queue>    
using namespace std;
typedef long long ll;
const int maxn = 100005;

class SAM
{
  const static int maxn = 400005;   //节点个数  
  const static int size = 11;     //字符的范围  
  const static char base = '0';     //字符的基准  

  class node
  {
  public:
    node *fa, *next[size];
    int len, sum, cnt, ins;
    node* clear(int x)
    {
      fa = 0; len = x;
      ins = sum = cnt = 0;
      memset(next, 0, sizeof(next));
      return this;
    }
  }nd[maxn];                      //节点的设置  

  node *root, *last;              //根节点,上一个节点  
  int tot;                        //总节点数  
public:
  void clear()
  {
    last = root = &nd[tot = 0];
    nd[0].clear(0);
  }                               //初始化  
  void insert(char ch)
  {
    node *p = last, *np = nd[++tot].clear(p->len + 1);
    last = np;
    int x = ch - base;
    while (p&&p->next[x] == 0) p->next[x] = np, np->ins++, p = p->fa;
    if (p == 0) { np->fa = root; return; }

    node* q = p->next[x];
    if (p->len + 1 == q->len) { np->fa = q; return; }

    node *nq = nd[++tot].clear(p->len + 1);
    for (int i = 0; i < size; i++) 
      if (q->next[i]) nq->next[i] = q->next[i], q->next[i]->ins++;
    nq->fa = q->fa;
    q->fa = np->fa = nq;
    while (p &&p->next[x] == q) p->next[x] = nq, nq->ins++, q->ins--, p = p->fa;
  }                               //插入操作  
  void query()
  {
    int ans = 0, base = 2012;
    root->cnt = 1;
    root->sum = 0;
    queue<node*> p;
    p.push(root);
    while (!p.empty())
    {
      node*q = p.front(); p.pop();
      (ans += q->sum) %= base;
      for (int i = 0; i < size; i++)
      if (q->next[i])
      {
        q->next[i]->ins--;
        if (q->next[i]->ins == 0) p.push(q->next[i]);
        if ((q == root&&i == 0) || i == 10) continue;
        (q->next[i]->cnt += q->cnt) %= base;
        (q->next[i]->sum += q->sum * 10 + q->cnt*i) %= base;
      }
    }
    printf("%d\n", ans);
  }
}sam;

int n;
char s[maxn];

int main()
{
  while (~scanf("%d", &n))
  {
    sam.clear();
    while (n--)
    {
      scanf("%s", s);
      for (int i = 0; s[i]; i++) sam.insert(s[i]);
      if (n) sam.insert('9' + 1);
    }
    sam.query();
  }
  return 0;
}