[TJOI2018]碱基序列

嘟嘟嘟




现在看到字符串就想到SAM，所以很担心kmp啥的会不会忘了……




这题感觉挺暴力的：首先当然要把\(s\)建成SAM，然后令\(dp[i][j]\)表示到第\(i\)组时，SAM上节点\(j\)能匹配的字符串个数。

转移的时候暴力枚举起点节点\(p\)，然后每一次都把当前字符串放上去跑，如果在SAM上存在的话，令结束节点为\(x\)，则有\(dp[i][x] += dp[i - 1][p]\)。

那么最后的答案就是\(\sum _ {i = 1} ^ {cnt} dp[m][i] * size[i]\)。

然后开一个临时数组就可以把dp降维了。

理论上\(O(|S| + ka_i |s|)\)的复杂度，实际上跑的飞快，挤到了loj rank3。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const ll mod = 1e9 + 7;
const int maxn = 1e4 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int m;
char s[maxn];

In ll inc(ll a, ll b) {return a + b >= mod ? a + b - mod : a + b;}

struct Sam
{
  int las, cnt;
  int tra[maxn << 1][26], len[maxn << 1], link[maxn << 1], siz[maxn << 1];
  In void init() {link[las = cnt = 0] = -1;}
  In void insert(int c)
  {
    int now = ++cnt, p = las;
    len[now] = len[las] + 1, siz[now] = 1;
    while(~p && !tra[p][c]) tra[p][c] = now, p = link[p];
    if(p == -1) link[now] = 0;
    else
      {
	int q = tra[p][c];
	if(len[q] == len[p] + 1) link[now] = q;
	else
	  {
	    int clo = ++cnt;
	    memcpy(tra[clo], tra[q], sizeof(tra[q]));
	    len[clo] = len[p] + 1;
	    link[clo] = link[q], link[q] = link[now] = clo;
	    while(~p && tra[p][c] == q) tra[p][c] = clo, p = link[p];
	  }
      }
    las = now;
  }
  char s[maxn];
  In int tour(int p, char* s)
  {
    int len = strlen(s);
    for(int i = 0, c; i < len; ++i)
      if(tra[p][c = s[i] - 'A']) p = tra[p][c];
      else return -1;
    return p;
  }
  int dp[maxn << 1] = {1}, tp[maxn << 1];
  In void solve()
  {
    fill(tp, tp + cnt + 1, 0);
    int T = read();
    while(T--)
      {
	scanf("%s", s);
	for(int i = 0; i <= cnt; ++i)
	  if(dp[i])
	    {
	      int v = tour(i, s);
	      if(~v) tp[v] = inc(tp[v], dp[i]);
	    }
      }
    for(int i = 0; i <= cnt; ++i) dp[i] = tp[i];
  }
  int buc[maxn << 1], pos[maxn << 1];
  In ll calc()
  {
    for(int i = 1; i <= cnt; ++i) ++buc[len[i]];
    for(int i = 1; i <= cnt; ++i) buc[i] += buc[i - 1];
    for(int i = 1; i <= cnt; ++i) pos[buc[len[i]]--] = i;
    ll ret = 0;
    for(int i = cnt; i; --i)
      {
	int now = pos[i], fa = link[now];
	siz[fa] += siz[now];
	ret = inc(ret, 1LL * dp[now] * siz[now] % mod);
      }
    return ret;
  }
}S;

int main()
{
  m = read(); scanf("%s", s);
  int len = strlen(s); S.init();
  for(int i = 0; i < len; ++i) S.insert(s[i] - 'A');
  for(int i = 1; i <= m; ++i) S.solve();
  write(S.calc());
  return 0;
}