UVa 11019 Matrix Matcher - Hash

题目传送门

　　快速的vjudge传送门

　　快速的UVa传送门

题目大意

　　给定两个矩阵S和T，问T在S中出现了多少次。

　　不会AC自动机做法。

　　考虑一维的字符串Hash怎么做。

　　对于一个长度为$l$的字符串$s$，它的Hash值$hash(s) = \sum_{i = 1}^{l}x^{l - i}s_{i}$。

　　对于二维的情况，我们就取两个基，$x, y$，对于一个$n\times m$的矩阵$A$的Hash值可以表示为

$hash(A) = \sum_{i = 1}^{n}\sum_{j = 1}^{m}x^{n - i}y^{m - j}a_{ij}$

　　然后以记录$S$的左上角的左上角的所有子矩阵的hash值（这个可以$O(1)$转移）。询问一个子矩阵的hash值，就可以$O(1)$回答。

　　接下来就很简单了。枚举每个位置判断是否匹配。

Code

 /**
  * UVa
  * Problem#11019
  * Accepted
  * Time: 50ms
  */
 #include <iostream>
 #include <cstdlib>
 #include <cstdio>
 using namespace std;
 typedef bool boolean;

 const unsigned int hash1 = , hash2 = ;
 const int N = , M = ;

 int p1[N], p2[N];
 int m, n, x, y;
 char S[N][N], T[M][M];
 unsigned int hs[N][N];

 inline void prepare() {
     p1[] = , p2[] = ;
     for (int i = ; i < N; i++)
         p1[i] = p1[i - ] * hash1;
     for (int i = ; i < N; i++)
         p2[i] = p2[i - ] * hash2;
 }

 inline void init() {
     scanf("%d%d", &n, &m);
     for (int i = ; i <= n; i++)
         scanf("%s", S[i] + );
     scanf("%d%d", &x, &y);
     for (int i = ; i <= x; i++)
         scanf("%s", T[i] + );
 }

 inline void solve() {
     for (int i = ; i <= n; i++)
         for (int j = ; j <= m; j++) {
             hs[i][j] = hs[i - ][j - ] * hash1 * hash2 + (hs[i - ][j] - hs[i - ][j - ] * hash2) * hash1 + (hs[i][j - ] - hs[i - ][j - ] * hash1) * hash2 + S[i][j];
         }

 /*    unsigned int s1 = 0;
     for (int i = 1; i <= n; i++)
         for (int j = 1; j <= m; j++)
             s1 += S[i][j] * p1[n - i] * p2[m - j];

     cerr << s1 << " " << (97u * 200379 * 211985 + 98u * 200379 + 98u * 211985 + 97) << " " << hs[2][2] << endl;*/

     int rt = ;
     unsigned int s = , c;
     for (int i = ; i <= x; i++)
         for (int j = ; j <= y; j++)
             s += T[i][j] * p1[x - i] * p2[y - j];
 //    cerr << s << endl;
     for (int i = x; i <= n; i++)
         for (int j = y; j <= m; j++) {
             c = hs[i][j] - hs[i - x][j - y] * p1[x] * p2[y] - (hs[i][j - y] - hs[i - x][j - y] * p1[x]) * p2[y] - (hs[i - x][j] - hs[i - x][j - y] * p2[y]) * p1[x];
             if (s == c)
                 rt++;
         }
     printf("%d\n", rt);
 }

 int kase;
 int main() {
     prepare();
     scanf("%d", &kase);
     while (kase--) {
         init();
         solve();
     }
     return ;
 }