3 seconds
256 megabytes
standard input
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).
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.
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.
4 1 1 1 1 2 1 1 0 1 1 1 0 1 0 0 1
12 2 2 3 2
#include<cstdio>
#include<cstring>
const int N = 2005;
typedef long long LL;
LL Map[N][N]; //棋盘
LL L[N*2]; //副对角线元素之和
LL R[N*2]; //主对角线元素之和
int main()
{
int n;
while(~scanf("%d",&n)) {
memset(L, 0, sizeof(L));
memset(R, 0, sizeof(R));
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= n; j++) {
scanf("%I64d",&Map[i][j]);
L[i+j] += Map[i][j]; //同一条副对角线i+j相同
R[i-j+n] += Map[i][j]; //同一条主对角线i-j相同
}
}
int x1 = 1, x2 = 1, y1 = 1, y2 = 2;
LL max1 = 0, max2 = 0;
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= n; j++) {
LL tmp = L[i+j] + R[i-j+n] - Map[i][j];
if((i+j) % 2 == 0 && tmp > max1) {
max1 = tmp;
x1 = i;
y1 = j;
}
if((i + j) % 2 == 1 && tmp > max2) {
max2 = tmp;
x2 = i;
y2 = j;
}
}
}
printf("%I64d\n", max1 + max2);
printf("%d %d %d %d\n", x1, y1, x2, y2);
}
return 0;
}