1. 题目信息

1.1 题目链接

http://poj.org/problem?id=3690

1.2 题目大意

给定大的矩阵（天空的样子），然后给定若干小矩阵（可能的天空的一角）
求有多少个小矩阵是从大矩阵里抠出来的（2D匹配）

1.3 解题思路

采用RK算法，求矩阵的哈希值，看哈希值是否一样，若一样，再比较一下看是否真的一样（防止哈希冲突）

2. 代码

2.1 Time Limit Exceeded 代码

计算的是整体矩阵的哈希值

/*** @description: poj3690 2维矩阵匹配* @author: michael ming* @date: 2019/6/25 19:47* @modified by: */
#include <iostream>
#include <math.h>
using namespace std;
int a[1001][1001];
int b[51][51];
typedef unsigned long long ull;
ull cal_hash_b(int r, int c, int b[][51])
{int i, j, k;ull value = 0;for (i = 0; i < r; ++i) //计算2d模式串的hash值value{for(j = 0, k = 1; j < c; ++j,++k)value += b[i][j]*pow(2.0,k);}return value;
}
ull cal_hash_a_child(int i0, int j0, int r, int c, int a[][1001])
{int i, j, k;ull hash_value = 0;for (i = i0; i < r; ++i) //计算2d子串的hash值value{for(j = j0, k = 1; j < c; ++j,++k)hash_value += a[i][j]*pow(2.0,k);}return hash_value;
}
bool same(int a[][1001], int b[][51], int i0, int j0, int mr, int mc)
{int x = i0, y = j0, i, j;for(i = 0; i < mr; ++i,++x){for(j = 0, y = j0; j < mc; ++j,++y)//记得写y=j0,换行后y复位{if(a[x][y] != b[i][j])return false;}}return true;
}
int str_RK_2d(int a[][1001], int nr, int nc, int b[][51], int mr, int mc)//s是主串，t是模式串
{int i, j;ull hash_val, value;value = cal_hash_b(mr,mc,b);//计算2d模式串哈希值for(i = 0; i < nr-mr+1; ++i)//行最多nr-mr+1次比较{for(j = 0; j < nc-mc+1; ++j)//列最多nc-mc+1次比较{hash_val = cal_hash_a_child(i,j,mr+i,mc+j,a);//计算2d子串哈希值if(hash_val == value && same(a,b,i,j,mr,mc)){//如果2d子串哈希值等于模式串的，且"真的"字符串匹配（避免冲突带来的假匹配）return 1;}}}return 0;
}
void creatMatrix_a(int a[][1001], int r, int c)
{int i, j;char ch;for(i = 0; i < r; ++i)for(j = 0; j < c; ++j){cin >> ch;if(ch == '*')a[i][j] = 1;elsea[i][j] = 0;}
}
void creatMatrix_b(int b[][51], int r, int c)
{int i, j;char ch;for(i = 0; i < r; ++i)for(j = 0; j < c; ++j){cin >> ch;if(ch == '*')b[i][j] = 1;elseb[i][j] = 0;}
}
int main()
{int N, M, T, P, Q, count, ID = 1;while(cin >> N >> M >> T >> P >> Q && N){count = 0;creatMatrix_a(a,N,M);while(T--){creatMatrix_b(b,P,Q);count += str_RK_2d(a,N,M,b,P,Q);}cout << "Case " << ID++ << ": " << count << endl;}return 0;
}

2.2 Time Limit Exceeded 代码

优化了哈希值的计算方式，采用错位乘以2的方式，2的k次幂提前算好(还试了改成位运算)，都是超时

/*** @description: poj3690 2维矩阵匹配* @author: michael ming* @date: 2019/6/25 19:47* @modified by:*/
#include <iostream>
using namespace std;
typedef unsigned long long ull;
int a[1001][1001];
int b[51][51];
ull cal_hash_b(int r, int c, int b[][51])
{int i, j, k;ull value = 0;for (i = 0; i < r; ++i) //计算2d模式串的hash值value{for(j = 0, k = c-1; j < c; ++j,--k){value += b[i][j]<<k;}}return value;
}
ull cal_hash_a_child(int i0, int j0, int r, int c, int a[][1001])
{int i, j, k;ull hash_value = 0;for (i = i0; i < r; ++i) //计算2d子串的hash值value{for(j = j0, k = c-1; j < c; ++j,--k)hash_value += a[i][j]<<k;}return hash_value;
}
bool same(int a[][1001], int b[][51], int i0, int j0, int mr, int mc)
{int x = i0, y = j0, i, j;for(i = 0; i < mr; ++i,++x){for(j = 0, y = j0; j < mc; ++j,++y)//记得写y=j0,换行后y复位{if(a[x][y] != b[i][j])return false;}}return true;
}
int sum(int a[][1001], int i0, int j0, int mr)
{int sum = 0;for(int x = 0; x < mr; ++x,++i0)sum += a[i0][j0];return sum;
}
int str_RK_2d(int a[][1001], int nr, int nc, int b[][51], int mr, int mc)//s是主串，t是模式串
{int i, j;ull hash_val, value;value = cal_hash_b(mr,mc,b);//计算2d模式串哈希值for(i = 0; i < nr-mr+1; ++i)//行最多nr-mr+1次比较{for(j = 0; j < nc-mc+1; ++j)//列最多nc-mc+1次比较{if(j == 0)hash_val = cal_hash_a_child(i,j,mr+i,mc+j,a);//计算2d子串哈希值elsehash_val = ((hash_val-(sum(a,i,j,mr)<<(mc-1)))<<1) + sum(a,i,j+mc-1,mr);if(hash_val == value && same(a,b,i,j,mr,mc)){//如果2d子串哈希值等于模式串的，且"真的"字符串匹配（避免冲突带来的假匹配）return 1;}}}return 0;
}
void creatMatrix_a(int a[][1001], int r, int c)
{int i, j;char ch;for(i = 0; i < r; ++i)for(j = 0; j < c; ++j){cin >> ch;if(ch == '*')a[i][j] = 1;elsea[i][j] = 0;}
}
void creatMatrix_b(int b[][51], int r, int c)
{int i, j;char ch;for(i = 0; i < r; ++i)for(j = 0; j < c; ++j){cin >> ch;if(ch == '*')b[i][j] = 1;elseb[i][j] = 0;}
}
int main()
{int N, M, T, P, Q, count, ID = 1;while(cin >> N >> M >> T >> P >> Q && N){count = 0;creatMatrix_a(a,N,M);while(T--){creatMatrix_b(b,P,Q);count += str_RK_2d(a,N,M,b,P,Q);}cout << "Case " << ID++ << ": " << count << endl;}return 0;
}

2.3 Time Limit Exceeded 代码

改为计算每行的哈希值，把大矩阵的所有小矩阵的宽度的每行哈希值算出来，后面开始逐行比较，有不符合的，跳出，寻找下一个，还是超时

/*** @description: poj3690 2维矩阵匹配* @author: michael ming* @date: 2019/6/25 19:47* @modified by:*/
#include <time.h>
#include <stdio.h>
typedef unsigned long long ull;
int a[1001][1001];
int b[51][51];
ull hash_b[51];//存放每行的哈希值，每行相当于一个2进制数
ull hash_a[1001][1001];//存放主串子串的哈希值void cal_hash_b(int r, int c, int b[][51])
{int i, j, k;ull value;for (i = 0; i < r; ++i) //计算2d模式串的hash值value{value = 0;for(j = 0, k = c-1; j < c; ++j,--k){value += b[i][j]<<k;}hash_b[i] = value;}return;
}
void cal_hash_a_child(int N, int M, int a[][1001], int P, int Q)
{int i, j, k, x;ull hash_value;for (i = 0; i < N; ++i) //计算2d子串的每行的hash值{for(j = 0; j < M-Q+1; ++j){if(j == 0){hash_value = 0;for(x = j, k = Q-1; x < j+Q && k >= 0; ++x,--k)hash_value += a[i][x]<<k;}elsehash_value = ((hash_a[i][j-1]-(a[i][j-1]<<(Q-1)))<<1)+a[i][j+Q-1];hash_a[i][j] = hash_value;}}
}int str_RK_2d(int a[][1001], int N, int M, int b[][51], int P, int Q)//s是主串，t是模式串
{int i, j, k, x;bool flag = false;cal_hash_b(P,Q,b);//计算2d模式串每行哈希值for(j = 0; j < M-Q+1; ++j)//列最多nc-mc+1次比较,分别比较每行，列先固定{for(i = 0; i < N-P+1; ++i){//行最多nr-mr+1次比较for(x = i, k = 0; x < i+P && k < P; ++x,++k){//一组比较P行if(hash_a[x][j] == hash_b[k])//比较子串哈希值flag = true;else{flag = false;break;}}if(flag == true)return 1;}}return 0;
}
void creatMatrix_a(int a[][1001], int r, int c)
{int i, j;char ch;for(i = 0; i < r; ++i)for(j = 0; j < c; ++j){ch = getchar();while(ch == ' '||ch == '\n')ch = getchar();if(ch == '*')a[i][j] = 1;elsea[i][j] = 0;}
}
void creatMatrix_b(int b[][51], int r, int c)
{int i, j;char ch;for(i = 0; i < r; ++i)for(j = 0; j < c; ++j){ch = getchar();while(ch == ' '||ch == '\n')ch = getchar();if(ch == '*')b[i][j] = 1;elseb[i][j] = 0;}
}
int main()
{
//    clock_t start, finish;
//    start = clock();int N, M, T, P, Q, count, ID = 1;while((scanf("%d%d%d%d%d",&N,&M,&T,&P,&Q)!=EOF) && N){count = 0;creatMatrix_a(a,N,M);cal_hash_a_child(N,M,a,P,Q);while(T--){creatMatrix_b(b,P,Q);count += str_RK_2d(a,N,M,b,P,Q);}printf("Case %d: %d\n",ID++,count);}
//    finish = clock();
//    cout << "takes "<< finish-start << " ms." << endl;return 0;
}