The Big Painting

题目连接：

Description

Samuel W. E. R. Craft is an artist with a growing reputation.

Unfortunately, the paintings he sells do not provide

him enough money for his daily expenses plus the new supplies

he needs. He had a brilliant idea yesterday when he

ran out of blank canvas: ”Why don’t I create a gigantic

new painting, made of all the unsellable paintings I have,

stitched together?”. After a full day of work, his masterpiece

was complete.

That’s when he received an unexpected phone call: a

client saw a photograph of one of his paintings and is willing

to buy it now! He had forgotten to tell the art gallery to

remove his old works from the catalog! He would usually

welcome a call like this, but how is he going to find his old

work in the huge figure in front of him?

Given a black-and-white representation of his original

painting and a black-and-white representation of his masterpiece, can you help S.W.E.R.C. identify in

how many locations his painting might be?

Input

The input file contains several test cases, each of them as described below.

The first line consists of 4 space-separated integers: hp wp hm wm, the height and width of the

painting he needs to find, and the height and width of his masterpiece, respectively.

The next hp lines have wp lower-case characters representing his painting. After that, the next hm

lines have wm lower-case characters representing his masterpiece. Each character will be either ‘x’ or

‘o’.

Constraints:

1 ≤ hp, wp ≤ 2 000

1 ≤ hm, wm ≤ 2 000

hp ≤ hm

wp ≤ wm

Output

The painting could be in four locations as shown in the following picture. Two of the locations overlap

Sample Input

4 4 10 10

oxxo

xoox

xoox

oxxo

xxxxxxoxxo

oxxoooxoox

xooxxxxoox

xooxxxoxxo

oxxoxxxxxx

ooooxxxxxx

xxxoxxoxxo

oooxooxoox

oooxooxoox

xxxoxxoxxo

Sample Output

4

Hint

题意

给你两个只含有ox的矩阵，问你第二个矩阵内有多少个第一个矩阵

题解：

标准的二维矩阵匹配，但是这道题数据范围太大了，ac自动机T成傻逼，不要问我为什么

所以就写hash吧。。。。

代码

#include 
    
     using namespace std;const long long pr=1e9+7;const int N=2005;const long long p=233LL;char s[N];long long hl[N*N],h[N][N],xh[N][N],a[N][N],b[N][N],tt;int main(){    int hp,wp,hm,wm;    while(scanf("%d%d%d%d",&hp,&wp,&hm,&wm)!=EOF)    {        memset(h,0,sizeof(h));        memset(xh,0,sizeof(xh));        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));    hl[0]=1LL;    for(int i=1;i<=wp*hp;i++)    {        hl[i]=(hl[i-1]*p)%pr;    }    tt=0;    for(int i=1;i<=hp;i++)    {        scanf("%s",s+1);        for(int j=1;j<=wp;j++)        {            if(s[j]=='o') a[i][j]=0;else a[i][j]=1;            tt=(tt*p+a[i][j])%pr;        }    }    for(int i=1;i<=hm;i++)    {        scanf("%s",s+1);        xh[i][0]=0;        for(int j=1;j<=wm;j++)        {            if(s[j]=='o') b[i][j]=0;else b[i][j]=1;            xh[i][j]=(xh[i][j-1]*p+b[i][j])%pr;        }    }    int cnt=0;    for(int j=1;j+wp-1<=wm;j++)    {        h[1][j]=0;        for(int k=1;k<=hp;k++)            h[1][j]=(h[1][j]*hl[wp]+(xh[k][j+wp-1]-xh[k][j-1]*hl[wp]%pr)+pr)%pr;        if(h[1][j]==tt) cnt++;        for(int i=2;i+hp-1<=hm;i++)        {            h[i][j]=((h[i-1][j]-(xh[i-1][j+wp-1]-xh[i-1][j-1]*hl[wp]%pr)*hl[wp*hp-wp]%pr)*hl[wp]%pr+                     (xh[i+hp-1][j+wp-1]-xh[i+hp-1][j-1]*hl[wp]%pr)+pr+pr)%pr;            if(h[i][j]==tt) cnt++;        }    }    cout<
     
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