【CODEFORCES】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 a_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), where x_i is the number of the row where the i-th bishop should be placed, y_i 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

题解：这一题看到题目数据不太大，首先想到枚举，然后竟然就这样写出来了......

2个相，不能同时攻击一个细胞，但是又要和最大，我们先处理出左对角线和又对角线的两个和，存在L[I]和R[I]中，然后分割棋盘 为  I+J为奇数的   和I+J为偶数的   这样两种情况分开取出最大值，最后记录输出即可。

分割棋盘这个方法太刁了....

注意要预处理X1 Y1 X2 Y2  不然最后会输出0 0 0 0 。（如果判断的时候是大于等于好像不用）

#include <iostream>
#include <cstdio>

using namespace std;

long long l[4002],r[4002];
long long  a[2002][2002],n,i,x2,x1,y2,y1,j;
int main()
{
    scanf("%I64d",&n);
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++) scanf("%I64d",&a[i][j]);
    //
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++)
    {
        l[n-j+i]+=a[i][j];
        r[i+j-1]+=a[i][j];
    }
    x1=1; y1=1;
    x2=1; y2=2;
    long long max1=0,max2=0;
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++)
        if ((i+j)%2!=0 && max1<l[n-j+i]+r[i+j-1]-a[i][j])
        {
            max1=l[n-j+i]+r[i+j-1]-a[i][j];
            x1=i;  y1=j;
        }
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++)
        if ((i+j)%2==0 && max2<l[n-j+i]+r[i+j-1]-a[i][j])
        {
            max2=l[n-j+i]+r[i+j-1]-a[i][j];
            x2=i; y2=j;
        }
    cout <<max1+max2<<endl;
    cout <<x1<<" "<<y1<<" "<<x2<<" "<<y2<<endl;
    return 0;
}