UVA - 11019 Matrix Matcher hash+KMP

题目链接：传送门

题解：

　　枚举每一行，每一行当中连续的y个我们hash 出来

　　那么一行就是 m - y + 1个hash值，形成的一个新 矩阵 大小是 n*(m - y + 1)， 我们要找到x*y这个矩阵的 话 是不是就是 在每一列跑kmp就行了？

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
typedef unsigned long long ULL;
const long long INF = 1e18+1LL;
const double pi = acos(-1.0);
const int N = 2e3+, M = 1e3+,inf = 2e9;

const ULL mod = 10000019ULL;
int n,m,x,y,T,fail[N];
char a[N][N],b[N][N];
ULL has[N][N],mp[N][N],s[N],t[N],sqr[N];
void build_fail() {
    int j = ;
    memset(fail,,sizeof(fail));
    for(int i = ; i <= y; ++i) {
        while(j&&s[i]!=s[j+]) j = fail[j];
        if(s[i] == s[j+]) j++;
        fail[i] = j;
    }
}
int kmp() {
    int ret = , j = ;
    for(int i = ; i <= n; ++i) {
        while(j&&s[j+]!=t[i]) j = fail[j];
        if(s[j+] == t[i]) j++;
        if(j == x) {
            ret += ;
            j = fail[j];
        }
    }
    return ret;
}
int main() {
    sqr[] = ;
    for(int i = ; i < N; ++i) sqr[i] = sqr[i-] * mod;
    scanf("%d",&T);
    while(T--) {
        scanf("%d%d",&n,&m);
        for(int i = ; i <= n; ++i) scanf("%s",a[i]+);
        scanf("%d%d",&x,&y);
        for(int i = ; i <= x; ++i) scanf("%s",b[i]+);
        if(n < x || m < y) {
            puts("");
            continue;
        }
        for(int i = ; i <= n; ++i) {
            for(int j = ; j <= m; ++j)
                has[i][j] = has[i][j - ] * mod + a[i][j];
        }
        for(int i = ; i <= x; ++i) {
            ULL now = , ps = ;
            for(int j = ; j <= y; ++j) {
                now = now * mod + b[i][j];
            }
            s[i] = now;
        }
        build_fail();
        int ans = ;
        for(int j = ; j <= m - y + ; ++j) {
            for(int i = ; i <= n; ++i) {
                int l = j, r = j + y - ;
                ULL now = has[i][r] - has[i][l-] * sqr[r - l + ];
                t[i] = now;
            }
            ans += kmp();
        }
        printf("%d\n",ans);
    }
    return ;
}