• 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;
    }
    
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  • 原文地址:https://www.cnblogs.com/SovietPower/p/9362295.html
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