BZOJ 1170 [Balkan2007]Cipher Hash

来源：互联网发布：大数据采集业务编辑：程序博客网时间：2024/05/21 14:47

题目大意：给定一个二维矩阵，求出现次数最多的a*b的子矩阵

二维Hash，只要记住横纵的BASE不能相同就可以，爱怎么搞怎么搞

一开始写的自然溢出结果OLE 以为是自然溢出被卡掉了于是写了双取模…… 结果还是OLE

最后发现尼玛这题读入坑爹……字符串里有空格不说，满满的不可见字符是咋回事……

记住不要用scanf读入……可以用gets，或者fread，注意要把一开始的回车过滤掉

getchar读进来全是错的不知道怎么回事……

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define M 1010#define BASE1 2591#define BASE2 2593#define MOD1 1000000007#define MOD2 1000000009using namespace std;typedef long long ll;int m,n,a,b,ans_time;char s[M][M];ll sum1[M][M],sum2[M][M],ans1,ans2,BASE1_a[3]={0,1,1},BASE2_b[3]={0,1,1};namespace Hash_Table{    struct list{        ll hash_value1,hash_value2;        int times;        list *next;        list(ll _,ll __,list *___):            hash_value1(_),hash_value2(__),times(0),next(___) {}    }*head[1001001];    inline int& Hash(ll hash1,ll hash2)    {        list *temp;        int x=hash1*hash2%1001001;        for(temp=head[x];temp;temp=temp->next)            if(temp->hash_value1==hash1&&temp->hash_value2==hash2)                return temp->times;        return head[x]=new list(hash1,hash2,head[x]),head[x]->times;    }}void Output(){    int i,j,k,l;    for(i=a;i<=m;i++)        for(j=b;j<=n;j++)        {            ll hash1=((sum1[i][j]                     -sum1[i-a][j]*BASE1_a[1]                     -sum1[i][j-b]*BASE2_b[1]                     +sum1[i-a][j-b]*BASE1_a[1]%MOD1*BASE2_b[1]                     )%MOD1+MOD1)%MOD1;            ll hash2=((sum2[i][j]                     -sum2[i-a][j]*BASE1_a[2]                     -sum2[i][j-b]*BASE2_b[2]                     +sum2[i-a][j-b]*BASE1_a[2]%MOD2*BASE2_b[2]                     )%MOD2+MOD2)%MOD2;            if(hash1==ans1&&hash2==ans2)            {                for(k=i-a+1;k<=i;k++,puts(""))                    for(l=j-b+1;l<=j;l++)                        putchar(s[k][l]);                return ;            }        }}void Find_Position(){    int i,j;    for(i=a;i<=m;i++)        for(j=b;j<=n;j++)        {            ll hash1=((sum1[i][j]                     -sum1[i-a][j]*BASE1_a[1]                     -sum1[i][j-b]*BASE2_b[1]                     +sum1[i-a][j-b]*BASE1_a[1]%MOD1*BASE2_b[1]                     )%MOD1+MOD1)%MOD1;            ll hash2=((sum2[i][j]                     -sum2[i-a][j]*BASE1_a[2]                     -sum2[i][j-b]*BASE2_b[2]                     +sum2[i-a][j-b]*BASE1_a[2]%MOD2*BASE2_b[2]                     )%MOD2+MOD2)%MOD2;            if(hash1==ans1&&hash2==ans2)                printf("%d %d\n",i-a+1,j-b+1);        }}int main(){    int i,j;    scanf("%d %d\n",&m,&n);    for(i=1;i<=m;i++)    {        fread(s[i]+1,n+1,1,stdin);        for(j=1;j<=n;j++)            sum1[i][j]=sum2[i][j]=s[i][j];    }    for(i=1;i<=m;i++)        for(j=1;j<=n;j++)            sum1[i][j]=(sum1[i][j]+sum1[i-1][j]*BASE1)%MOD1,            sum2[i][j]=(sum2[i][j]+sum2[i-1][j]*BASE1)%MOD2;    for(i=1;i<=m;i++)        for(j=1;j<=n;j++)            sum1[i][j]=(sum1[i][j]+sum1[i][j-1]*BASE2)%MOD1,            sum2[i][j]=(sum2[i][j]+sum2[i][j-1]*BASE2)%MOD2;    cin>>a>>b;    cout<<a<<' '<<b<<endl;    for(i=1;i<=a;i++)    {        BASE1_a[1]*=BASE1;BASE1_a[1]%=MOD1;        BASE1_a[2]*=BASE1;BASE1_a[2]%=MOD2;    }    for(j=1;j<=b;j++)    {        BASE2_b[1]*=BASE2;BASE2_b[1]%=MOD1;        BASE2_b[2]*=BASE2;BASE2_b[2]%=MOD2;    }    for(i=a;i<=m;i++)        for(j=b;j<=n;j++)        {            ll hash1=((sum1[i][j]                     -sum1[i-a][j]*BASE1_a[1]                     -sum1[i][j-b]*BASE2_b[1]                     +sum1[i-a][j-b]*BASE1_a[1]%MOD1*BASE2_b[1]                     )%MOD1+MOD1)%MOD1;            ll hash2=((sum2[i][j]                     -sum2[i-a][j]*BASE1_a[2]                     -sum2[i][j-b]*BASE2_b[2]                     +sum2[i-a][j-b]*BASE1_a[2]%MOD2*BASE2_b[2]                     )%MOD2+MOD2)%MOD2;            int& val=Hash_Table::Hash(hash1,hash2);            if(++val>ans_time)                ans_time=val,ans1=hash1,ans2=hash2;        }    Output();    cout<<ans_time<<endl;    Find_Position();    return 0;}

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