POJ3690 Constellations

嘟嘟嘟


哈希


刚开始我一直在想二维哈希，但发现如果还是按行列枚举的话会破坏子矩阵的性质。也就是说，这个哈希只能维护一维的子区间的哈希值。
所以我就开了个二维数组\(has_{i, j}\)表示原矩阵\(s_{i, j - q + 1}\)到\(s_{i, j}\)的哈希值，所以这个要用滚动哈希。
滚动哈希就是这样的：\(hash[s_{i, i + m}] = hash[s_{i + 1, j + m + 1}] * base - s_i * base ^ m\)。理解起来就是把\(s_i\)对哈希的贡献减去。
然后用同样的方法算出\(p * q\)矩阵的哈希值。最后逐行比对。
时间复杂度瓶颈在于比对复杂度\(O(n ^ 2 * p)\)。

#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 rg register
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
const ull bas = 19260817;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e3 + 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 n, m, t, p, q;

char tp[maxn];
ull has[maxn][maxn], a[maxn];

bool judge()
{
  for(int i = 1; i <= n - p + 1; ++i)
    for(int j = q; j <= m; ++j)
      {
	bool flg = 1;
	for(int k = 1; k <= p && flg; ++k)
	  if(has[i + k - 1][j] != a[k]) flg = 0;
	if(flg) return 1;
      }
  return 0;
}

ull quickpow(ull a, int b)
{
  ull ret = 1;
  for(; b; b >>= 1, a *= a)
    if(b & 1) ret *= a;
  return ret;
}

int main()
{
  int tcnt = 0;
  while(scanf("%d", &n) && n)
    {
      m = read(); t = read(); p = read(); q = read();
      for(int i = 1; i <= n; ++i)
	{
	  scanf("%s", tp + 1);
	  for(int j = 1; j <= m; ++j)
	    {
	      has[i][j] = has[i][j - 1] * bas + tp[j];
	      if(j >= q + 1) has[i][j] -= quickpow(bas, q) * tp[j - q];
	    }
	}
      int ans = 0;
      while(t--)
	{
	  for(int i = 1; i <= p; ++i)
	    {
	      a[i] = 0;
	      scanf("%s", tp + 1);
	      for(int j = 1; j <= q; ++j) a[i] = a[i] * bas + tp[j];
	    }
	  if(judge()) ans++;
	}
      printf("Case %d: %d\n", ++tcnt, ans);
    }
  return 0;
}