CF463C——Gargari and Bishops(模擬)

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 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.

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

題意：

就是在一個棋盤上擺放兩個國際象棋中的象，象能夠攻擊到與它在同一條斜線上的對象。

現在給你一個N*N的棋盤，其中每個格子都有個值，代表攻擊後的得分。

要求你求出擺放兩個象所能獲得的最大的分數。並且這兩個象的攻擊範圍不能有重合。

分析：

直接看代碼的註釋吧。

可以注意到：

同一條從左上到右下的斜線上的元素滿足 i-j  相同。

同一條從左下到右上的斜線上的元素滿足 i+j 相同。

#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <iostream>
#define INF 0x7fffffff
using namespace std;

typedef long long LL;
const int N = 2e3 + 10;

int mat[N][N];
LL L[3*N]; //從左下到右上的斜線上的元素和
LL R[3*N]; //從右下到左上的斜線上的元素和
LL G[N][N]; //在改點擺放棋子所能獲得的分數

int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0; i<n; i++)       //讀入元素，預處理L,R
        for(int j=0; j<n; j++){
            scanf("%d",&mat[i][j]);
            L[i+j]+=mat[i][j];
            R[i-j+n]+=mat[i][j];
        }

    for(int i=0;i<n;i++){       //計算G[][]
        for(int j=0;j<n;j++){
            G[i][j]=L[i+j]+R[i-j+n]-mat[i][j];
        }
    }

    int x1,x2,y1,y2;
    x1=x2=y1=y2=0;             
    for(int i=0;i<n;i++){      //貪心求第一個棋子的擺放位置 
        for(int j=0;j<n;j++){
            if(G[x1][y1]<=G[i][j])
                x1=i,y1=j;
        }
    }
    for(int i=0;i<n;i++){     //求第二個棋子擺放的位置
        for(int j=0;j<n;j++){
            if((x1+y1)%2==(i+j)%2)  //兩個棋子不能有重合的攻擊範圍
                continue;
            if(G[x2][y2]<=G[i][j])
                x2=i,y2=j;
        }
    }

    LL sum=G[x1][y1]+G[x2][y2];

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

    return 0;
}