UVA 11019 Matrix Matcher （hash+kmp）

11019 Matrix Matcher

Given an N × M matrix, your task is to find the number of occurences of an X × Y pattern.

Input

The first line contains a single integer t (t ≤ 15), the number of test cases.For each case, the first line contains two integers N and M (N, M ≤ 1000). The next N linescontain M characters each.The next line contains two integers X and Y (X, Y ≤ 100). The next X lines contain Y characterseach.

Output

For each case, output a single integer in its own line, the number of occurrences.

Sample Input

21 1x1 1y3 3abcbcdcde2 2bccd

Sample Output

02

解题思路：大矩阵（m*n）小矩阵（x*y）将大矩阵每一行hash，大矩阵每一列储存包括他本身在内 y 列的信息，小矩阵也按照这种方式压缩成 x*1 的矩阵，然后将大矩阵存储的那些数值按照一列一列的储存进一个数组中，每一列之间用一个不可能出现的数字分割，然后在大矩阵的hash数组中跑小矩阵hash数组的kmp算法。

AC代码：

/*    @Author: wchhlbt    @Date:   2017/8/7*///#include <bits/stdc++.h>#include <vector>#include <list>#include <map>#include <set>#include <queue>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <ctime>#include <cstring>#include <limits>#include <climits>#include <cstdio>#define Fori(x) for(int i=0;i<x;i++)#define Forj(x) for(int j=0;j<x;j++)#define maxn 1003#define inf 0x3f3f3f3f#define ONES(x) __builtin_popcount(x)#define _  << "  " <<using namespace std;typedef long long ll ;const double eps =1e-8;const int mod = 1000000007;typedef pair<int, int> P;const double PI = acos(-1.0);int dx[4] = {0,0,1,-1};int dy[4] = {1,-1,0,0};inline int read(){ int num;    scanf("%d",&num);   return num;}int n,m;int ans;/* * 把字符串变成X进制数，可以完成$ O(1) $比较 , * 调用 Init() 初始化幂 , * 调用 Init(u64* Hash, int len) 初始化 Hash 数组 , * Get(u64* Hash, int p, int L) 表示获得以 p 开头长度为 L 的字符串 Hash 。 * Base 需要选用质数 */typedef unsigned long long u64;const u64 Base = 31;const int N = 1003;char s[N][N];u64 Hash[N];u64 Pow[N];void Init() {    Pow[0] = 1;    for(int i = 1; i < N; i++) Pow[i] = Pow[i - 1] * Base;}void Init(int id, int len) {    Hash[len] = 0;    for(int i = len - 1; i >= 0; i--)        Hash[i] = (u64)Hash[i + 1] * Base + (s[id][i] - 'a' + 1);}u64 Get(int p, int L) {    return Hash[p] - Hash[p + L] * Pow[L];}u64 dic[maxn][maxn];//KMPu64 p[10003];u64 a[1000007];int f[maxn];//失配函数void getFail(int plen){    f[0] = 0; f[1] = 0;    for(int i = 1; i<plen; i++){        int j = f[i];        while(j && p[i]!=p[j])  j = f[j];        f[i+1] = (p[i]==p[j]) ? j+1 : 0;    }}int find(int slen, int plen)//s->待匹配文本串  p->模板串{    getFail(plen);    int j = 0;    int cnt = 0;    for(int i = 0; i<slen; i++){        while(j && a[i]!=p[j])  j = f[j];        if(a[i]==p[j])            j++;        if(j==plen){            j = f[j];//不考虑重叠时，此处置0            //return i + 1 -plen; 返回第一个匹配的位置            cnt++;        }    }    return cnt;}int main(){    //freopen("test.txt","w",stdout);    int t = read();    Init();    while(t--){        int n = read(); int m = read();        int cnt = 0;                for(int i = 1; i<=n; i++){            scanf("%s",s[i]);                    }        int x = read(); int y = read();        int slen;        for(int i = 1; i<=n; i++){            slen = strlen(s[i]);            Init(i,slen);            for(int j = 0; j+y<=slen; j++)                dic[i][j] = Get(j,y);        }        for(int i = 1; i<=x; i++){            scanf("%s",s[i]);            int len = strlen(s[i]);            Init(i,len);            p[cnt++] = Hash[0];        }        int plen = cnt;        cnt = 0;        for(int j = 0; j+y<=slen; j++){            for(int i = 1; i<=n; i++){                a[cnt++] = dic[i][j];            }            a[cnt++] = -1;        }        int alen = cnt;        printf("%d\n", find(alen,plen));    }    return 0;}/*unsigned   int   0～4294967295int   2147483648～2147483647unsigned long 0～4294967295long   2147483648～2147483647long long的最大值：9223372036854775807long long的最小值：-9223372036854775808unsigned long long的最大值：18446744073709551615__int64的最大值：9223372036854775807__int64的最小值：-9223372036854775808unsigned __int64的最大值：18446744073709551615*/