BZOJ.2462.[BeiJing2011]矩阵模板(二维Hash)

题目链接

序列上的Hash和前缀和差不多，二维Hash也和二维前缀和差不多了。

预处理大矩阵所有r*c的小矩阵hash值，再对询问的矩阵Hash。

类比于序列上\(s[r]-s[l-1]*pow[r-l+1]\)，比如\(s[i-r][j-c]\)多算了\(r*c\)次，乘个\(pow[r]*pow[c]\)就行。

用指针替掉数组的一维竟然慢点。。好吧毕竟也就一维，数据也不大。

//106864kb	1520ms
#include <cstdio>
#include <cctype>
#include <algorithm>
//#define gc() getchar()
#define MAXIN 500000
#define gc() (SS==TT&&(TT=(SS=IN)+fread(IN,1,MAXIN,stdin),SS==TT)?EOF:*SS++)
#define P 100000000
#define base1 12289
#define base2 786433
typedef unsigned int uint;//在这用unsigned long long好像没太大用吧...果然，光荣地T了
const int N=1005;
uint A[N][N],sum[N][N],pw1[N],pw2[N];
bool Hash[P];
char IN[MAXIN],*SS=IN,*TT=IN;
inline int read()
{
	int now=0;register char c=gc();
	for(;!isdigit(c);c=gc());
	for(;isdigit(c);now=now*10+c-'0',c=gc());
	return now;
}
inline int read01()
{
	register char c=gc();
	for(;!isdigit(c);c=gc());
	return c-'0';
}
int main()
{
	int n=read(),m=read(),r=read(),c=read();
	for(int i=1; i<=n; ++i)
		for(int j=1; j<=m; ++j) sum[i][j]=read01();
	pw1[0]=pw2[0]=1;
	for(int i=1; i<=n; ++i) pw1[i]=pw1[i-1]*base1;
	for(int i=1; i<=m; ++i) pw2[i]=pw2[i-1]*base2;
	for(int i=2; i<=n; ++i)//Row
		for(int j=1; j<=m; ++j) sum[i][j]+=sum[i-1][j]*base1;
	for(int i=1; i<=n; ++i)//Column
		for(int j=2; j<=m; ++j) sum[i][j]+=sum[i][j-1]*base2;
	for(int i=r; i<=n; ++i)
		for(int j=c; j<=m; ++j)
		{
			uint hash=sum[i][j]-sum[i-r][j]*pw1[r]-sum[i][j-c]*pw2[c]+sum[i-r][j-c]*pw1[r]
*pw2[c];
			Hash[hash%P]=1;
		}
	for(int Q=read(); Q--; )
	{
		for(int i=1; i<=r; ++i)
			for(int j=1; j<=c; ++j) A[i][j]=read01();
		for(int i=2; i<=r; ++i)
			for(int j=1; j<=c; ++j) A[i][j]+=A[i-1][j]*base1;
		for(int i=1; i<=r; ++i)
			for(int j=2; j<=c; ++j) A[i][j]+=A[i][j-1]*base2;
		puts(Hash[A[r][c]%P]?"1":"0");
	}
	return 0;
}

//106864kb	1544ms
#include <cstdio>
#include <cctype>
#include <algorithm>
//#define gc() getchar()
#define MAXIN 500000
#define gc() (SS==TT&&(TT=(SS=IN)+fread(IN,1,MAXIN,stdin),SS==TT)?EOF:*SS++)
#define P 100000000
#define base1 12289
#define base2 786433
typedef unsigned int uint;//在这用unsigned long long好像没太大用吧... 果然，光荣地T了
const int N=1005;
uint A[N][N],sum[N][N],pw1[N],pw2[N];
bool Hash[P];
char IN[MAXIN],*SS=IN,*TT=IN;
inline int read()
{
	int now=0;register char c=gc();
	for(;!isdigit(c);c=gc());
	for(;isdigit(c);now=now*10+c-'0',c=gc());
	return now;
}
inline int read01()
{
	register char c=gc();
	for(;!isdigit(c);c=gc());
	return c-'0';
}
int main()
{
	int n=read(),m=read(),r=read(),c=read();
	for(int i=1; i<=n; ++i)
		for(int j=1; j<=m; ++j) sum[i][j]=read01();
	pw1[0]=pw2[0]=1;
	for(int i=1; i<=n; ++i) pw1[i]=pw1[i-1]*base1;
	for(int i=1; i<=m; ++i) pw2[i]=pw2[i-1]*base2;
	for(int i=2; i<=n; ++i)//Row
	{
		uint *s=sum[i], *las=sum[i-1];
		for(int j=1; j<=m; ++j) s[j]+=las[j]*base1;
	}
	for(int i=1; i<=n; ++i)//Column
	{
		uint *s=sum[i];
		for(int j=2; j<=m; ++j) s[j]+=s[j-1]*base2;
	}
	for(int i=r; i<=n; ++i)
		for(int j=c; j<=m; ++j)
		{
			uint hash=sum[i][j]-sum[i-r][j]*pw1[r]-sum[i][j-c]*pw2[c]+sum[i-r][j-c]*pw1[r]*pw2[c];
			Hash[hash%P]=1;
		}
	for(int Q=read(); Q--; )
	{
		for(int i=1; i<=r; ++i)
			for(int j=1; j<=c; ++j) A[i][j]=read01();
		for(int i=2; i<=r; ++i)
		{
			uint *s=A[i], *las=A[i-1];
			for(int j=1; j<=c; ++j) s[j]+=las[j]*base1;
		}
		for(int i=1; i<=r; ++i)
		{
			uint *s=A[i];
			for(int j=2; j<=c; ++j) s[j]+=s[j-1]*base2;
		}
		puts(Hash[A[r][c]%P]?"1":"0");
	}
	return 0;
}