【CODEFORCES】C. 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 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
41 1 1 12 1 1 01 1 1 01 0 0 1
output
122 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;}

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