UVA11019 Matrix Matcher【hash傻逼题】【AC自动机好题】

LINK1

LINK2

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题目大意

让你在一个大小为\(n*m\)的矩阵中找大小是\(x*y\)的矩阵的出现次数

思路1：Hash

hash思路及其傻逼

你把一维情况扩展一下

一维是一个bas，那你二维就用两个bas好了

对一个在\((i,j)\)的字符，令他的hash值是\(c_{i,j}*bas1^i*bas2^j\)

然后算出矩阵hash值乘上差量判断就做完了

70ms

---

//Author: dream_maker
#include<bits/stdc++.h>
using namespace std;
//----------------------------------------------
typedef pair<int, int> pi;
typedef long long ll;
typedef double db;
#define fi first
#define se second
#define fu(a, b, c) for (int a = b; a <= c; ++a)
#define fd(a, b, c) for (int a = b; a >= c; --a)
#define fv(a, b) for (int a = 0; a < (signed)b.size(); ++a)
const int INF_of_int = 1e9;
const ll INF_of_ll = 1e18;
template <typename T>
void Read(T &x) {
  bool w = 1;x = 0;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-') w = 0, c = getchar();
  while (isdigit(c)) {
    x = (x<<1) + (x<<3) + c -'0';
    c = getchar();
  }
  if (!w) x = -x;
}
template <typename T>
void Write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) Write(x / 10);
  putchar(x % 10 + '0');
}
//----------------------------------------------
const int N = 1e3 + 10;
const int Mod = 998244353;
const int bas1 = 233333;
const int bas2 = 19260817;
int pow1[N], pow2[N];
int n, m, x, y;
char s[N][N], c[N][N];
int sum[N][N], val;
int add(int a, int b) {
  return (a += b) >= Mod ? a - Mod : a;
}
int mul(int a, int b) {
  return 1ll * a * b % Mod;
}
int sub(int a, int b) {
  return (a -= b) < 0 ? a + Mod : a;
}
int fast_pow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    b >>= 1;
    a = mul(a, a);
  }
  return res;
}
void init() {
  pow1[0] = pow2[0] = 1;
  fu(i, 1, N - 1) pow1[i] = mul(bas1, pow1[i - 1]);
  fu(i, 1, N - 1) pow2[i] = mul(bas2, pow2[i - 1]);
}
void getsum() {
  fu(i, 1, n)
    fu(j, 1, m)
      sum[i][j] = add(sub(add(sum[i][j - 1], sum[i - 1][j]), sum[i - 1][j - 1]), mul(s[i][j], mul(pow1[i], pow2[j])));
  val = 0;
  fu(i, 1, x)
    fu(j, 1, y) val = add(val, mul(c[i][j], mul(pow1[i], pow2[j])));
}
void solve() {
  Read(n), Read(m);
  fu(i, 1, n) scanf("%s", s[i] + 1);
  Read(x), Read(y);
  fu(i, 1, x) scanf("%s", c[i] + 1);
  getsum();
  int ans = 0;
  fu(i, x, n)
    fu(j, y, m)
      if (sub(add(sum[i][j], sum[i - x][j - y]), add(sum[i][j - y], sum[i - x][j])) == mul(val, mul(pow1[i - x], pow2[j - y])))
        ++ans;
  Write(ans), putchar('\n');
}
int main() {
#ifdef dream_maker
  freopen("input.txt", "r", stdin);
#endif
  int T; Read(T);
  init();
  while (T--) solve();
  return 0;
}

---

思路2：AC自动机

用AC自动机来考虑的话这题挺好的

虽然跑600ms

考虑一下把模式串分解变成x个长度是y的串

然后全部塞进AC自动机

然后考虑算出在\(n*m\)的矩阵中有哪些串在哪些位置出现过

这个东西跑一边就可以处理出来

如果有不好处理的细节你就想怎么暴力怎么来

然后我们考虑假如在\((i,j)\)这个位置匹配到了第k行

那么对于左上角在\((i-k,j)\)的矩阵显然是可以匹配第k行的

那么我们就记录一下每个节点是左上角的矩阵最多能匹配多少行就可以了

---

//Author: dream_maker
#include<bits/stdc++.h>
using namespace std;
//----------------------------------------------
typedef pair<int, int> pi;
typedef long long ll;
typedef double db;
#define fi first
#define se second
#define fu(a, b, c) for (int a = b; a <= c; ++a)
#define fd(a, b, c) for (int a = b; a >= c; --a)
#define fv(a, b) for (int a = 0; a < (signed)b.size(); ++a)
const int INF_of_int = 1e9;
const ll INF_of_ll = 1e18;
template <typename T>
void Read(T &x) {
  bool w = 1;x = 0;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-') w = 0, c = getchar();
  while (isdigit(c)) {
    x = (x<<1) + (x<<3) + c -'0';
    c = getchar();
  }
  if (!w) x = -x;
}
template <typename T>
void Write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) Write(x / 10);
  putchar(x % 10 + '0');
}
//----------------------------------------------
const int N = 1e3 + 10;
const int CHARSET_SIZE = 26;
struct Node {
  int ch[CHARSET_SIZE], fail;
  int id[110];
  Node() {}
  void clean() {
    fu(i, 0, CHARSET_SIZE - 1) ch[i] = 0;
    id[0] = fail = 0;
  }
} p[N * N];
int n, m, x, y, cnt;
int res[N][N];
char s[N][N], c[N][N];
void init() {
  cnt = 1;
  p[0].clean(); p[1].clean();
  fu(i, 0, CHARSET_SIZE - 1) p[0].ch[i] = 1;
}
void insert(char *c, int id) {
  int u = 1;
  fu(i, 1, y) {
    int tmp = c[i] - 'a';
    if (!p[u].ch[tmp])
      p[p[u].ch[tmp] = ++cnt].clean();
    u = p[u].ch[tmp];
  }
  p[u].id[++p[u].id[0]] = id;
}
void build_fail() {
  static queue<int> q;
  q.push(1);
  while (q.size()) {
    int u = q.front(); q.pop();
    fu(i, 0, CHARSET_SIZE - 1) {
      int w = p[u].ch[i], v = p[u].fail;
      while (!p[v].ch[i]) v = p[v].fail;
      v = p[v].ch[i];
      if (w) {
        p[w].fail = v;
        q.push(w);
      } else p[u].ch[i] = v;
    }
  }
}
void trans(char *c, int id) {
  int u = 1;
  fu(i, 1, m) {
    u = p[u].ch[c[i] - 'a'];
    fu(j, 1, p[u].id[0]) {
      if (id >= p[u].id[j])
        res[id - p[u].id[j] + 1][i]++;
    }
  }
}
void solve() {
  init();
  Read(n), Read(m);
  fu(i, 1, n) scanf("%s", s[i] + 1);
  Read(x), Read(y);
  fu(i, 1, x) {
    scanf("%s", c[i] + 1);
    insert(c[i], i);
  }
  build_fail();
  fu(i, 1, n) fu(j, 1, m) res[i][j] = 0;
  fu(i, 1, n) trans(s[i], i);
  int ans = 0;
  fu(i, 1, n)
    fu(j, 1, m) if (res[i][j] == x) ++ans;
  Write(ans), putchar('\n');
}
int main() {
#ifdef dream_maker
  freopen("input.txt", "r", stdin);
#endif
  int T; Read(T);
  while (T--) solve();
  return 0;
}