【BZOJ 2351】 Matrix

【题目链接】

            https://www.lydsy.com/JudgeOnline/problem.php?id=2351

【算法】

            哈希

【代码】

             

#include<bits/stdc++.h>
using namespace std;
#define MAXN 1010
typedef unsigned long long ull;
const int base1 = 131;
const int base2 = 211;
const int P = 1e5 + 7;

int i,j,m,n,a,b,q;
ull sum[MAXN][MAXN];
ull power1[MAXN],power2[MAXN];
ull t;
vector< ull > e[P];
char c;
char s[MAXN][MAXN];

inline void insert(ull x)
{
        int h = (int)(x % P);
        e[h].push_back(x);        
}
inline bool query(ull x)
{
        int i;
        int h = (int)(x % P);
        for (i = 0; i < (int)e[h].size(); i++)
        {
                if (e[h][i] == x)
                        return true;
        }
        return false;
}
 
int main() 
{
        
        scanf("%d%d%d%d",&m,&n,&a,&b);
        for (i = 1; i <= m; i++) scanf("%s",s[i]+1);
        for (i = 1; i <= m; i++)
        {
                for (j = 1; j <= n; j++)
                {
                        sum[i][j] = s[i][j] - '0';
                }
        }
        for (i = 1; i <= m; i++)
        {
                for (j = 1; j <= n; j++)
                {
                        sum[i][j] += sum[i-1][j] * base1;
                }
        }
        for (i = 1; i <= m; i++)
        {
                for (j = 1; j <= n; j++)
                {
                        sum[i][j] += sum[i][j-1] * base2;
                }
        }
        power1[0] = power2[0] = 1;
        for (i = 1; i < m; i++)
        {
                power1[i] = power1[i-1] * base1;
                power2[i] = power2[i-1] * base2;
        }
        for (i = a; i <= m; i++)
        {
                for (j = b; j <= n; j++)
                {
                        t = sum[i][j] - sum[i-a][j] * power1[a] - sum[i][j-b] * power2[b] + sum[i-a][j-b] * power1[a] * power2[b];
                        insert(t); 
                }
        }
        scanf("%d",&q);
        while (q--)
        {
                for (i = 1; i <= a; i++) scanf("%s",s[i]+1);
                for (i = 1; i <= a; i++)
                {
                        for (j = 1; j <= b; j++)
                        {
                                sum[i][j] = s[i][j] - '0';
                        }
                }
                for (i = 1; i <= a; i++)
                {
                        for (j = 1; j <= b; j++)
                        {
                                sum[i][j] += sum[i-1][j] * base1;
                        }
                }
                for (i = 1; i <= a; i++)
                {
                        for (j = 1; j <= b; j++)
                        {
                                sum[i][j] += sum[i][j-1] * base2;
                        }
                }
                if (query(sum[a][b])) printf("1\n");
                else printf("0\n");
        }
        
        return 0;
    
}

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转载自www.cnblogs.com/evenbao/p/9282310.html
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