codeforces 173C Spiral Maximum(简单预处理)

C. Spiral Maximum

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Let's consider a k × k square, divided into unit squares. Please note that k ≥ 3 and is odd. We'll paint squares starting from the upper left square in the following order: first we move to the right, then down, then to the left, then up, then to the right again and so on. We finish moving in some direction in one of two cases: either we've reached the square's border or the square following after the next square is already painted. We finish painting at the moment when we cannot move in any direction and paint a square. The figure that consists of the painted squares is a spiral.

The figure shows examples of spirals for k = 3, 5, 7, 9.

You have an n × m table, each of its cells contains a number. Let's consider all possible spirals, formed by the table cells. It means that we consider all spirals of any size that don't go beyond the borders of the table. Let's find the sum of the numbers of the cells that form the spiral. You have to find the maximum of those values among all spirals.

Input

The first line contains two integers n and m (3 ≤ n, m ≤ 500) — the sizes of the table.

Each of the next n lines contains m space-separated integers: the j-th number in the i-th line a_ij ( - 1000 ≤ a_ij ≤ 1000) is the number recorded in the j-th cell of the i-th row of the table.

Output

Print a single number — the maximum sum of numbers among all spirals.

Sample test(s)

input

6 5
0 0 0 0 0
1 1 1 1 1
0 0 0 0 1
1 1 1 0 1
1 0 0 0 1
1 1 1 1 1

output

17

input

3 3
1 1 1
1 0 0
1 1 1

output

6

input

6 6
-3 2 0 1 5 -1
4 -1 2 -3 0 1
-5 1 2 4 1 -2
0 -2 1 3 -1 2
3 1 4 -3 -2 0
-1 2 -1 3 1 2

output

13

Note

In the first sample the spiral with maximum sum will cover all 1's of the table.

In the second sample the spiral may cover only six 1's.

题意：

给你一个n * m (3 ≤ n, m ≤ 500)的矩阵。定义了一种螺旋矩阵。矩阵的边长为>=3的奇数。矩阵的值为矩阵的螺旋线上格子的权值和。螺旋线的定义为。从矩阵的左上角开始。依次按左，下，右，上的方向前进。转向的条件是当前格子的当前方向的下下个格子被走过了或到了矩阵的边界。结束的条件是不能再前进了。现在告诉你权值矩阵。要你找出子矩阵的最大权值。

思路：

为了减少频繁的求和操作我们预处理出rsum[i][j],第i行的前j项和。csum[j][i]。第j列的前i项和。通过观察会发现n=7比n=3.多出的部分和n=9比n=5多出的部分都是外围的不到一圈加一点。所以n=3,7,11,15的可以一圈一圈的加。n=5,9,13的和也可以一圈一圈的加。所以我们只需要枚举子矩阵的左上角和长度就可以算出最大权值了。最大时间复杂度为(500*500*250).这是最坏的估计。实际比这好的多。

详细见代码：

#include<bits/stdc++.h>
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=1e-8;
const double PI=acos(-1.0);
const int maxn=100010;
int rsum[510][510],csum[510][510],val[510][510];

int main()
{
    int n,m,i,j,ans,sum,x1,y1,x2,y2;

    while(~scanf("%d%d",&n,&m))
    {
        for(i=1;i<=n;i++)
            for(j=1;j<=m;j++)
                scanf("%d",&val[i][j]);
        for(i=1;i<=n;i++)
        {
            rsum[i][0]=0;
            for(j=1;j<=m;j++)
                rsum[i][j]=rsum[i][j-1]+val[i][j];
        }
        for(j=1;j<=m;j++)
        {
            csum[j][0]=0;
            for(i=1;i<=n;i++)
                csum[j][i]=csum[j][i-1]+val[i][j];
        }
        ans=-INF;
        for(i=1;i<=n;i++)
            for(j=1;j<=m;j++)
            {
                x1=i,y1=j;
                x2=x1+2,y2=y1+2;
                if(x2<=n&&y2<=m)//3,7,11,15....
                {
                    sum=rsum[x1][y2]-rsum[x1][y1-1]+csum[y2][x2]-csum[y2][x1]+rsum[x2][y2-1]-rsum[x2][y1-1];
                    ans=max(ans,sum);
                    while(x1-2>=1&&y1-2>=1&&x2+2<=n&&y2+2<=m)
                    {
                        x1-=2,y1-=2,x2+=2,y2+=2;
                        sum+=rsum[x1][y2]-rsum[x1][y1-1]+csum[y2][x2]-csum[y2][x1]+rsum[x2][y2-1]-rsum[x2][y1];
                        sum+=csum[y1][x2]-csum[y1][x1+1]+val[x1+2][y1+1];
                        ans=max(ans,sum);
                    }
                }
                x1=i,y1=j;
                x2=x1+4,y2=y1+4;
                if(x2<=n&&y2<=m)//5,9,13,17.....
                {
                    sum=rsum[x1][y2]-rsum[x1][y1-1]+csum[y2][x2]-csum[y2][x1]+rsum[x2][y2-1]-rsum[x2][y1];
                    sum+=csum[y1][x2]-csum[y1][x1+1]+val[x1+2][y1+1]+val[x1+2][y1+2];
                    ans=max(ans,sum);
                    while(x1-2>=1&&y1-2>=1&&x2+2<=n&&y2+2<=m)
                    {
                        x1-=2,y1-=2,x2+=2,y2+=2;
                        sum+=rsum[x1][y2]-rsum[x1][y1-1]+csum[y2][x2]-csum[y2][x1]+rsum[x2][y2-1]-rsum[x2][y1];
                        sum+=csum[y1][x2]-csum[y1][x1+1]+val[x1+2][y1+1];
                        ans=max(ans,sum);
                    }
                }
            }
        printf("%d\n",ans);
    }
    return 0;
}