LA 6893 The Big Painting（矩阵Hash）

https://vjudge.net/problem/UVALive-6893

题意：

给出一个小矩阵和大矩阵，在大矩阵中能找到相同的小矩阵。

思路：

矩阵Hash，先对小矩阵计算出它的Hash值，然后处理大矩阵，计算出每个子矩阵的Hash值，然后和小矩阵的Hash值比较是否相等。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
#include<queue>
#include<cmath>
using namespace std;

typedef unsigned long long ULL;

const ULL Base1 = 2007;
const ULL Base2 = 2009;

int test,n,m,x,y;
ULL ans;
char s[2005][2005],a[2005][2005];
ULL hash;
ULL temp[2005][2005],Temp[2005][2005];

ULL Gethash()
{
    ULL c,sum = 0;
    for(int i = 0; i < x; ++i)
    {
        c = 0;
        for(int j = 0; j < y; ++j)
        {
            c = c*Base1 + a[i][j];
        }
        sum = sum * Base2 + c;
    }
    return sum;
}

//计算出每个子矩阵的Hash值，并比较Hash值是否相等
void GetAns()
{
    ULL t=1, c;
    for(int i = 0; i < y; ++i) t *= Base1;
    for(int i = 0; i < n; ++i)
    {
        c = 0;
        for(int j = 0; j < y; ++j) c = c*Base1 + s[i][j];
        temp[i][y-1] = c;
        for(int j = y; j < m; ++j)
        {
            temp[i][j]=temp[i][j-1]*Base1-s[i][j-y]*t+s[i][j];
        }
    }

    t = 1;
    for(int i = 0; i < x; ++i) t *= Base2;
    for(int i = y-1; i < m; ++i)
    {
        c = 0;
        for(int j = 0; j < x; ++j) c = c*Base2 + temp[j][i];
        Temp[x-1][i] = c;
        if(c == hash) ans++;
        for(int j = x; j < n; ++j)
        {
            Temp[j][i] = Temp[j-1][i]*Base2 - temp[j-x][i] * t + temp[j][i];
            if(Temp[j][i]== hash) ans++;
        }
    }
}

int main()
{
    //freopen("D:\input.txt","r",stdin);
    while(scanf("%d%d%d%d",&x,&y,&n,&m)!= EOF)
    {
        for(int i = 0; i < x; ++i) scanf("%s",a[i]);
        for(int i = 0; i < n; ++i) scanf("%s",s[i]);
        hash = Gethash();
        ans = 0;
        GetAns();
        printf("%llu
",ans);
    }
    return 0;
}