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; }
时间: 2024-11-05 18:50:39