学步园

题目大意：给出一个m*n的由01组成的矩阵，下面有q个询问，查询矩阵中存不存在大小为k*l的子矩阵。

思路：二维hash。我们先把大矩阵hash，然后把所有可能的k*l的子矩阵都插到哈希表里，然后只要对于每个询问hash一下看哈希表中是否存在。

值得一提的是，这个题只需要输出10个1就可以AC了。。

CODE：

#include <cstdio>
#include <bitset>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 1100
using namespace std;
const unsigned int BASE1 = 10016957;
const unsigned int BASE2 = 10016957;
const int MO = 99999997;
 
int m,n,ask_m,ask_n,asks;
unsigned int hash[MAX][MAX],_hash[MAX][MAX];
unsigned int pow1[MAX],pow2[MAX];
 
bool set[100000000];
 
inline unsigned int GetHash()
{
    for(int i = 1; i <= ask_m; ++i)
        for(int j = 1; j <= ask_n; ++j)
            _hash[i][j] += _hash[i - 1][j] * BASE1;
    for(int i = 1; i <= ask_m; ++i)
        for(int j = 1; j <= ask_n; ++j)
            _hash[i][j] += _hash[i][j - 1] * BASE2;
    return _hash[ask_m][ask_n];
}
 
int main()
{
    cin >> m >> n >> ask_m >> ask_n;
    for(int i = 1; i <= m; ++i)
        for(int j = 1; j <= n; ++j)
            scanf("%1d",&hash[i][j]);
    pow1[0] = pow2[0] = 1;
    for(int i = 1; i <= 100; ++i)
        pow1[i] = pow1[i - 1] * BASE1,pow2[i] = pow2[i - 1] * BASE2;
    for(int i = 1; i <= m; ++i)
        for(int j = 1; j <= n; ++j)
            hash[i][j] += hash[i - 1][j] * BASE1;
    for(int i = 1; i <= m; ++i)
        for(int j = 1; j <= n; ++j)
            hash[i][j] += hash[i][j - 1] * BASE2;
    for(int i = ask_m; i <= m; ++i)
        for(int j = ask_n; j <= n; ++j) {
            unsigned int h = hash[i][j];
            h -= hash[i - ask_m][j] * pow1[ask_m];
            h -= hash[i][j - ask_n] * pow2[ask_n];
            h += hash[i - ask_m][j - ask_n] * pow1[ask_m] * pow2[ask_n];
            set[h % MO] = true;
        }
    for(cin >> asks; asks--;) {
        for(int i = 1; i <= ask_m; ++i)
            for(int j = 1; j <= ask_n; ++j)
                scanf("%1d",&_hash[i][j]);
        puts(set[GetHash() % MO] ? "1":"0");
    }
    return 0;
}