[CF677E] Vanya and Balloons

题目大意

给你一个只有0,1,2的n*n的矩阵，你现在可以选择一个叉叉或者十字，把里面的数全部取掉，贡献为取掉的数的乘积。
具体的，从某一点(i,j)出发，选择一个长度c:
你可以这样取(1为取，0为不取)
00100
00100
11111
00100
00100
或者这样取
10001
01010
00100
01010
10001
n<=1000

分析

很简单的思路，弄两种，共4个前缀和数组，维护两个对角线的前缀和，还有一行一列的前缀和，然后枚举以哪个点作为中心。然后怎么确定这个点怎么取呢？我们知道如果要取，在取不到0时，肯定c越大越好；取到0就没有意义了。所以两种取法，分别二分长度，看看最大的c是多少，用log比较大小，更新答案即可。
其实原来我还怕精度不够，想直接弄double比较，然而精度其实差不多，而且double只有301位，即 21000 2^{1000}，如果全是3的话早爆了。
这道题比较麻烦的地方是对角线前缀和数组，要码得快必须要先规定好表示法，然后再打，这样才能思路清晰，不至于打得慢。
其实这题最好用直角坐标系，然后直接设数组bx[i][j]表示y=±x+i这条直线，会比较好做。
然而我没有转成坐标系，直接bx[i][j]表示起点为(i,1)，到(i+j-1,j)的前缀和。另一条对角线类似，以(i,n)为起点，为了储存地址都是正数，还要整体移n。

代码

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;
typedef double db;
#define fo(i,j,k) for(i=j;i<=k;i++)
#define fd(i,j,k) for(i=j;i>=k;i--)
const int N=2005;
const ll mo=1e9+7;
struct rec
{int c[4];
}ax[N][N],ay[N][N],bx[N][N],by[N][N],ret,ans;
int i,j,k,n,m,l,r,mid,a[N][N],xx,x,y,mn,z,prt;
db two[N];
char ch;
rec ins(rec a,int z)
{a.c[z]++;return a;
}
bool operator <(rec a,rec b)
{if (a.c[0]) return 1;if (b.c[0]) return 0;db aa,bb;aa=log(2)*(db)a.c[2]+log(3)*(db)a.c[3];bb=log(2)*(db)b.c[2]+log(3)*(db)b.c[3];return aa<bb;
}
rec operator +(rec a,rec b)
{a.c[0]=a.c[0]+b.c[0];a.c[1]=a.c[1]+b.c[1];a.c[2]=a.c[2]+b.c[2];a.c[3]=a.c[3]+b.c[3];return a;
}
rec operator -(rec a,rec b)
{a.c[0]=a.c[0]-b.c[0];a.c[1]=a.c[1]-b.c[1];a.c[2]=a.c[2]-b.c[2];a.c[3]=a.c[3]-b.c[3];return a;
}
void get1(int x)
{ret=ax[i][j+x]-ax[i][j-x-1]+ay[j][i+x]-ay[j][i-x-1];
}
void get2(int x)
{int y=n-j+1;ret=bx[i-j+n+1][j+x]-bx[i-j+n+1][j-x-1]+by[i+j][y+x]-by[i+j][y-x-1];
}
int main()
{
//  freopen("677.in","r",stdin);scanf("%d\n",&n);fo(i,1,n) {fo(j,1,n){scanf("%c",&ch);a[i][j]=ch-'0';}scanf("\n");}ans.c[0]=1;fo(i,1,n)fo(j,1,n){ax[i][j]=ins(ax[i][j-1],a[i][j]);ay[j][i]=ins(ay[j][i-1],a[i][j]);}fo(xx,-n+2,n){i=xx+n;x=xx;z=n;fo(y,1,n){if (x>0){bx[i][y]=ins(bx[i][y-1],a[x][y]);by[i][y]=ins(by[i][y-1],a[x][z]);}x++;z--;}}fo(i,1,n)fo(j,1,n)if (a[i][j]){// non-rotatedmn=min(min(i-1,n-i),min(j-1,n-j));l=0;r=mn;while (l<r){mid=(l+r+1)/2;get1(mid);if (ret.c[0]==0) l=mid;else r=mid-1;}get1(l);ret.c[a[i][j]]--;if (ans<ret)ans=ret;//rotatedl=0;r=mn;while (l<r){mid=(l+r+1)/2;get2(mid);if (ret.c[0]==0) l=mid;else r=mid-1;}get2(l);ret.c[a[i][j]]--;if (ans<ret)ans=ret;}prt=1;fo(i,0,3)fo(j,1,ans.c[i])prt=(ll)prt*i%mo;printf("%d\n",prt);
}

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[CF677E] Vanya and Balloons

题目大意

分析

代码

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