CodeForces 463C - Gargari and Bishops

Gargari is jealous that his friend Caisa won the game from the previousproblem. 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 thereis no cell that is attacked by both of them. Consider a cell with number x written
on it, if this cell is attacked by oneof the bishops Gargari will get x dollars for it. Tell Gargari, how to
place bishops on the chessboard toget maximum amount of money.

We assume a cell is attacked by a bishop, if the cell is located on thesame diagonal with the bishop (the cell, where the bishop is, also consideredattacked 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 thechessboard.

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 thenumber 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 numberedfrom 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

思路：

用f1,f2两个数组分别存储反对角线，正对角线上的数的和：

f1[i + j] += a[i][j];

f2[i - j + n] += a[i][j];

然后点(i,j)上两条对角线的和便为

sum[i][j] = f1[i + j] + f2[i - j + n] - a[i][j];

比赛时候没看到Gargari wants to place two bishops on thechessboard in such a way that there is no cell that is attacked by both ofthem.即两只棋子对角线不会有交集-.-

代码：
#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int N = 2005;

__int64 sum

;
__int64 f1[N * 2], f2[N * 2];
int a[N * 2][N * 2];

int main()
{
int n;

memset(f1, 0, sizeof(f1));
memset(f2, 0, sizeof(f2));

scanf("%d", &n);
int i;
for (i = 0; i<n; i++)
{
int j;
for (j = 0; j<n; j++)
{
scanf("%d", &a[i][j]);
f1[i + j] += a[i][j];
f2[i - j + n] += a[i][j];
}
}
for (i = 0; i<n; i++)
for (int j = 0; j<n; j++)
sum[i][j] = f1[i + j] + f2[i - j + n] - a[i][j];

int x1, y1, x2, y2;
__int64 max1 = 0, max2 = 0;
x1 = x2 = y1 = 1;
y2 = 2;
for (i = 0; i<n; i++)
for (int j = 0; j<n; j++)
{
if ((i + j) & 1)
{
if (max1<sum[i][j])
{
max1 = sum[i][j];
x1 = i;
y1 = j;
}
}
else
{
if (max2<sum[i][j])
{
max2 = sum[i][j];
x2 = i;
y2 = j;
}
}
}

printf("%I64d\n", max1 + max2);
printf("%d %d %d %d\n", x1 + 1, y1 + 1, x2 + 1, y2 + 1);

return 0;
}