Algorithm

C. Gargari and Bishops
time limit per test
3 seconds
memory limit per test
256 megabytes
problem link- https://codeforces.com/contest/463/problem/C
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.

Examples
input
Copy
4
1 1 1 1
2 1 1 0
1 1 1 0
1 0 0 1
output
Copy
12
2 2 3 2



 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include <bits/stdc++.h>

using namespace std;

int const N = 3001;
int n, g[N][N], x, y, xx, yy;
long long dp1[2 * N], dp2[2 * N], sol1 = -1, sol2 = -1;

int main() {
  scanf("%d", &n);
  for(int i = 1; i <= n; ++i)
    for(int j = 1; j <= n; ++j) {
      scanf("%d", &g[i][j]);
      dp1[i + j] += g[i][j];
      dp2[i - j + n] += g[i][j];
    }
  
  for(int i = 1; i <= n; ++i)
    for(int j = 1; j <= n; ++j) {
      long long v = dp1[i + j] + dp2[i - j + n] - g[i][j];
      if((i + j) & 1) {
        if(v > sol2) {
          sol2 = v;
          xx = i;
          yy = j;
        }
      } else {
        if(v > sol1) {
          sol1 = v;
          x = i;
          y = j;
        }
      }
    }
  
  printf("%lld\n", sol1 + sol2);
  printf("%d %d %d %d\n", x, y, xx, yy);
  
  return 0;
}
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Input

x
–
+
cmd
4
1 1 1 1
2 1 1 0
1 1 1 0
1 0 0 1

Output

x
–
+
cmd
12
2 2 3 2
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Demonstration

Codeforces Solution-Gargari and Bishops-Solution in C, C++, Java, Python

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