bzoj 1414: [ZJOI2009]对称的正方形

来源：互联网发布：淘宝的心怎么算得编辑：程序博客网时间：2024/05/19 19:15

题意：

求多少个二维的回文串。

题解：

向四个方向hash，枚举中间点，分奇偶讨论，二分边长。
unsigned int AC unsigned long long WA是什么鬼。
code：

#include<cstdio>#include<cstdlib>#include<cstring>#include<iostream>#define LL unsigned intusing namespace std;const LL base1=233,base2=2333;LL pre1[1001],pre2[1001];int n,m;LL hash[4][1001][1001];int a[1010][1001];int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}    return x*f;}LL get0(int x,int y,int x1,int y1){return hash[0][x1][y1]-hash[0][x1][y-1]*pre1[y1-y+1]-hash[0][x-1][y1]*pre2[x1-x+1]+hash[0][x-1][y-1]*pre2[x1-x+1]*pre1[y1-y+1];}LL get2(int x,int y,int x1,int y1){return hash[2][x][y]-hash[2][x1+1][y]*pre2[x1-x+1]-hash[2][x][y1+1]*pre1[y1-y+1]+hash[2][x1+1][y1+1]*pre2[x1-x+1]*pre1[y1-y+1];}LL get1(int x,int y,int x1,int y1){return hash[1][x1][y]-hash[1][x1][y1+1]*pre1[y1-y+1]-hash[1][x-1][y]*pre2[x1-x+1]+hash[1][x-1][y1+1]*pre2[x1-x+1]*pre1[y1-y+1];}LL get3(int x,int y,int x1,int y1){return hash[3][x][y1]-hash[3][x][y-1]*pre1[y1-y+1]-hash[3][x1+1][y1]*pre2[x1-x+1]+hash[3][x1+1][y-1]*pre2[x1-x+1]*pre1[y1-y+1];}int main(){    n=read();m=read();    for(int i=1;i<=n;i++)        for(int j=1;j<=m;j++) a[i][j]=read();    pre1[0]=pre2[0]=1;    for(int i=1;i<=1000;i++) pre1[i]=pre1[i-1]*base1,pre2[i]=pre2[i-1]*base2;    for(int j=1;j<=m;j++)        for(int i=1;i<=n;i++) hash[0][i][j]=hash[0][i-1][j]*base2+a[i][j];    for(int j=1;j<=m;j++)        for(int i=1;i<=n;i++) hash[1][i][j]=hash[1][i-1][j]*base2+a[i][j];    for(int j=1;j<=m;j++)        for(int i=n;i>=1;i--) hash[2][i][j]=hash[2][i+1][j]*base2+a[i][j];    for(int j=1;j<=m;j++)        for(int i=n;i>=1;i--) hash[3][i][j]=hash[3][i+1][j]*base2+a[i][j];    for(int i=1;i<=n;i++)        for(int j=1;j<=m;j++) hash[0][i][j]+=hash[0][i][j-1]*base1;    for(int i=1;i<=n;i++)        for(int j=1;j<=m;j++) hash[3][i][j]+=hash[3][i][j-1]*base1;    for(int i=1;i<=n;i++)        for(int j=m;j>=1;j--) hash[1][i][j]+=hash[1][i][j+1]*base1;    for(int i=1;i<=n;i++)        for(int j=m;j>=1;j--) hash[2][i][j]+=hash[2][i][j+1]*base1;    int ans=0;    for(int i=1;i<=n;i++)        for(int j=1;j<=m;j++)        {            int l=1,r=min(min(i,n-i+1),min(j,m-j+1)),t=1;            while(l<=r)            {                int mid=(l+r)/2;                LL n0=get0(i-mid+1,j-mid+1,i,j);                LL n1=get1(i-mid+1,j,i,j+mid-1);                LL n2=get2(i,j,i+mid-1,j+mid-1);                LL n3=get3(i,j-mid+1,i+mid-1,j);                if(n0==n1&&n1==n2&&n2==n3) l=mid+1,t=mid;                else r=mid-1;            }            ans+=t;        }    for(int i=1;i<n;i++)        for(int j=1;j<m;j++)        {            int l=1,r=min(min(i,n-i),min(j,m-j)),t=0;            while(l<=r)            {                int mid=(l+r)/2;                LL n0=get0(i-mid+1,j-mid+1,i,j);                LL n1=get1(i-mid+1,j+1,i,j+mid);                LL n2=get2(i+1,j+1,i+mid,j+mid);                LL n3=get3(i+1,j-mid+1,i+mid,j);                if(n0==n1&&n1==n2&&n2==n3) l=mid+1,t=mid;                else r=mid-1;            }            ans+=t;        }    printf("%d",ans);}

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