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 = ;
 const ULL Base2 = ;

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

 ULL Gethash()
 {
     ULL c,sum = ;
     for(int i = ; i < x; ++i)
     {
         c = ;
         for(int j = ; 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=, c;
     for(int i = ; i < y; ++i) t *= Base1;
     for(int i = ; i < n; ++i)
     {
         c = ;
         for(int j = ; j < y; ++j) c = c*Base1 + s[i][j];
         temp[i][y-] = c;
         for(int j = y; j < m; ++j)
         {
             temp[i][j]=temp[i][j-]*Base1-s[i][j-y]*t+s[i][j];
         }
     }

     t = ;
     for(int i = ; i < x; ++i) t *= Base2;
     for(int i = y-; i < m; ++i)
     {
         c = ;
         for(int j = ; j < x; ++j) c = c*Base2 + temp[j][i];
         Temp[x-][i] = c;
         if(c == hash) ans++;
         for(int j = x; j < n; ++j)
         {
             Temp[j][i] = Temp[j-][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 = ; i < x; ++i) scanf("%s",a[i]);
         for(int i = ; i < n; ++i) scanf("%s",s[i]);
         hash = Gethash();
         ans = ;
         GetAns();
         printf("%llu\n",ans);
     }
     return ;
 }