CF463C——Gargari and Bishops(模拟)

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

41 1 1 12 1 1 01 1 1 01 0 0 1

output

122 2 3 2

题意：

就是在一个棋盘上摆放两个国际象棋中的象，象能够攻击到与它在同一条斜线上的对象。

现在给你一个N*N的棋盘，其中每个格子都有个值，代表攻击后的得分。

要求你求出摆放两个象所能获得的最大的分数。并且这两个象的攻击范围不能有重合。

分析：

直接看代码的注释吧。

可以注意到：

同一条从左上到右下的斜线上的元素满足 i-j  相同。

同一条从左下到右上的斜线上的元素满足 i+j 相同。

#include <algorithm>#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <iostream>#define INF 0x7fffffffusing namespace std;typedef long long LL;const int N = 2e3 + 10;int mat[N][N];LL L[3*N]; //从左下到右上的斜线上的元素和LL R[3*N]; //从右下到左上的斜线上的元素和LL G[N][N]; //在改点摆放棋子所能获得的分数int main(){    int n;    scanf("%d",&n);    for(int i=0; i<n; i++)       //读入元素，预处理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++){       //计算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++){      //贪心求第一个棋子的摆放位置         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++){     //求第二个棋子摆放的位置        for(int j=0;j<n;j++){            if((x1+y1)%2==(i+j)%2)  //两个棋子不能有重合的攻击范围                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;}