Problem H
Matrix Matcher
Input: Standard Input

Output: Standard Output

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 lines contain M characters each.

The next line contains two integers X and Y (X, Y ≤ 100). The next X lines contain Y characters each.

Output

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

Sample Input Output for Sample Input

2

1 1

x

1 1

y

3 3

abc

bcd

cde

2 2

bc

cd

0

2

Problem Setter: Rujia Liu, EPS

Special Thanks: Wenbin Tang

Warming: The judge input file size is about 7 MB. So please make sure that you use a fast IO function (eg.scanf()) to read input.

每次成功匹配，就把以當前行匹配算出來的矩陣的右上角cnt++，被+x次說明匹配矩陣成功。

注意vector中的元素用(*it)

PS：由於P中單詞長度一致，故不用在print循環last

PS2:有可能同一行出現2次覆蓋v，因此要用vector

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define MAXT (15+10)
#define MAXN (1000+10)
#define MAXX (100+10)
#define MAXNode (1000000)
#define Sigma_size (26)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}

int cnt[MAXN][MAXN]; //右上角 

class Aho_Corasick_Automata
{
public:
	int ch[MAXNode][Sigma_size],siz; 
	vector v[MAXNode];
	// AC自動機 
	int f[MAXNode],last[MAXNode];
	Aho_Corasick_Automata(int _siz=0):siz(_siz){MEM(ch) Rep(i,MAXNode) v[i].clear(); MEM(f) MEM(last)}
	void mem(int _siz=0){siz=_siz; MEM(ch) Rep(i,MAXNode) v[i].clear(); MEM(f) MEM(last)	}
	int idx(char c){return c-'a';}
	void insert(char *s,int val=1) //val!=0 表示單詞末尾，       PS:lrj叫它單詞節點  
	{
		int u=0,n=strlen(s);
		Rep(i,n)
		{
			int c=idx(s[i]);
			if (!ch[u][c])
			{
				++siz;
				MEM(ch[siz]);
				ch[u][c]=siz;
			}
			u=ch[u][c];
		}
		v[u].push_back(val);
	}
	void getFail()
	{
		queue q;
		Rep(c,Sigma_size)
		{
			int u=ch[0][c];
			if (u) q.push(u),last[u]=0;
		}
		while (!q.empty())
		{
			int r=q.front();q.pop();  //r--c-->u
			Rep(c,Sigma_size)
			{
				int u=ch[r][c];
				if (!u) {ch[r][c]=ch[f[r]][c]; continue;} 
				q.push(u);
				f[u]=ch[f[r]][c];
				last[u]=v[f[u]].size()?f[u]:last[f[u]];
			}
		}
	} 
	void print(int j,int r,int c) //打印全串中所有以j為末尾的str 
	{
		for(vector::iterator it=v[j].begin();it!=v[j].end();it++)
		{
			int P_i=(*it);
			if (r-(P_i-1)<0) continue ; 
			cnt[r-(P_i-1)][c]++;
		}
	} 
	void find(char *s,int r)
	{
		int u=0,n=strlen(s);
		Rep(i,n)
		{
			int c=idx(s[i]);
			u=ch[u][c];
			if (v[u].size()) print(u,r,i);
			else if (last[u]) print(u,r,i);
		}
	}
	
}T;
int n,m,x,y;
char s[MAXN][MAXN];
char s2[MAXN];
int main()
{
//	freopen("uva11019.in","r",stdin);
//	freopen(".out","w",stdout);
	int tt;
	scanf("%d",&tt);
	while(tt--)
	{
		T.mem();
		scanf("%d%d",&n,&m);
		Rep(i,n)
		{
			scanf("%s",s[i]);
		}
		
		scanf("%d%d",&x,&y);
		For(i,x)
		{
			scanf("%s",s2);
			T.insert(s2,i);
		}
		T.getFail();
		
		MEM(cnt)
		
		Rep(i,n)
		{
			char *str=s[i];
			T.find(str,i);
		}
		int ans=0;
		Rep(i,n)
		{
			Rep(j,m)
				ans+=(bool)(cnt[i][j]==x);
		}
		cout<