BZOJ 2462 BeiJing 2011 矩阵模板 二维hash

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 1100using 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;}