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 = 200379, hash2 = 211985;
const int N = 1005, M = 105;

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[0] = 1, p2[0] = 1;
    for (int i = 1; i < N; i++)
        p1[i] = p1[i - 1] * hash1;
    for (int i = 1; i < N; i++)
        p2[i] = p2[i - 1] * hash2;
}

inline void init() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
        scanf("%s", S[i] + 1);
    scanf("%d%d", &x, &y);
    for (int i = 1; i <= x; i++)
        scanf("%s", T[i] + 1);
}

inline void solve() {
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++) {
            hs[i][j] = hs[i - 1][j - 1] * hash1 * hash2 + (hs[i - 1][j] - hs[i - 1][j - 1] * hash2) * hash1 + (hs[i][j - 1] - hs[i - 1][j - 1] * 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 = 0;
    unsigned int s = 0, c;
    for (int i = 1; i <= x; i++)
        for (int j = 1; 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 0;
}