codeforces C#264. Gargari and Bishops

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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 getx 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 integersa_ij(0 ≤ a_ij ≤ 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), wherex_i is the number of the row where thei-th bishop should be placed, y_i is the number of the column where thei-th bishop should be placed. Consider rows are numbered from 1 ton from top to bottom, and columns are numbered from 1 ton from left to right.

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

Sample test(s)

Input

Output

12
2 2 3 2
题意：给你个n*n的矩阵，对于一个点我们可以统计经过它的斜线上的值，让你选两个点，且经过的斜线是不相交的，求这两个点能组成的最大权值
思路：首先预处理出主负斜线上的值，然后就是把矩阵划分成类似国际象棋的黑白格子，分别计算黑白格子的最大值
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
typedef long long ll;
using namespace std;
const int maxn = 2005;ll arr[maxn*2];
ll brr[maxn*2];
int num[maxn][maxn];int main() {int n, a;scanf("%d", &n);memset(arr, 0, sizeof(arr));memset(brr, 0, sizeof(brr));for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) {scanf("%d", &num[i][j]);}for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++) {arr[i-j+n] += num[i][j];	brr[i+j] += num[i][j];}int x1, y1, x2, y2;ll Max1 = -1, Max2 = -1;for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++) {if (abs(i+j) % 2 == 0) {if (Max1 < arr[i-j+n] + brr[i+j] - num[i][j]) {Max1 = arr[i-j+n] + brr[i+j] - num[i][j];x1 = i, y1 = j;}}if (abs(i+j) % 2 == 1) {if (Max2 < arr[i-j+n] + brr[i+j] - num[i][j]) {Max2 = arr[i-j+n] + brr[i+j] - num[i][j];x2 = i, y2 = j;}}}cout << Max1+Max2 << endl;printf("%d %d %d %d\n", x1, y1, x2, y2);return 0;
}

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