Codeforces Round 264(div2) C. Gargari and Bishops

C. Gargari and Bishops

time limit per test
3 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Gargari is jealous that his friend Caisa won the game from the previous problem. He wants to prove that he is a genius.

He has a n × n chessboard. Each cell of the chessboard has a number written on it. Gargari wants to place two bishops on the chessboard in such
a way that there is no cell that is attacked by both of them. Consider a cell with number x written on it, if this cell is attacked by one of the bishops
Gargari will get x dollars for it. Tell Gargari, how to place bishops on the chessboard to get maximum amount of money.

We assume a cell is attacked by a bishop, if the cell is located on the same diagonal with the bishop (the cell, where the bishop is, also considered attacked by it).

Input

The first line contains a single integer n (2 ≤ n ≤ 2000).
Each of the next n lines contains n integers aij (0 ≤ aij ≤ 109) —
description of the chessboard.

Output

On the first line print the maximal number of dollars Gargari will get. On the next line print four integers: x1, y1, x2, y2 (1 ≤ x1, y1, x2, y2 ≤ n),
where xi is
the number of the row where the i-th bishop should be placed, yi is
the number of the column where the i-th bishop should be placed. Consider rows are numbered from 1 to n from
top to bottom, and columns are numbered from 1 to n from left to right.

If there are several optimal solutions, you can print any of them.

Sample test(s)

input

4
1 1 1 1
2 1 1 0
1 1 1 0
1 0 0 1

output

12
2 2 3 2

在半个小时的时候，思路就出来了，后来实现时自己把自己绕进去了，悲哀

，其实特别简单，斜着的判断，左上到右下的i+j相同，右上到左下的i-j相同，还有重点的判断，其实两个点i+j的奇偶性不同就行了。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn=2000+100;
int a[maxn][maxn];
long long hash1[2*maxn+10];
long long hash2[2*maxn+10];
queue<int> xz;
queue<int> yz;
int main()
{
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
scanf("%d",&a[i][j]);
}
}
memset(hash1,0,sizeof(hash1));
memset(hash2,0,sizeof(hash2));
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
hash1[i+j]+=a[i][j];
hash2[i-j+maxn]+=a[i][j];
}
}
long long ans,max1=0,max2=0;
int x1=1,y1=1;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if((hash1[i+j]+hash2[i-j+maxn]-a[i][j])>max1)
{
max1=hash1[i+j]+hash2[i-j+maxn]-a[i][j];
x1=i;
y1=j;
}
}
}
int x2=1;
int y2=2;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if((i+j)%2==!((x1+y1)%2))
{
if(hash1[i+j]+hash2[i-j+maxn]-a[i][j]>max2)
{
max2=hash1[i+j]+hash2[i-j+maxn]-a[i][j];
x2=i;
y2=j;
}
}
}
}
ans=max1+max2;
printf("%I64d\n",ans);
printf("%d %d %d %d\n",x1,y1,x2,y2);
return 0;
}

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