bzoj 1414: [ZJOI2009]对称的正方形
2017-10-10 08:33
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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 int using 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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