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).
InputThe 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.
Sample test(s) input4 1 1 1 1 2 1 1 0 1 1 1 0 1 0 0 1output
12 2 2 3 2题意：给出一个n*n的棋盘，每个方格中有一个非负整数，在棋盘中放两个象，使得这两个象既不会互相攻击，攻击范围也不能有重合，并且在这两个象攻击范围内的格子中的值的总和最大，输出总和和两个象的位置坐标。如果一个格子和象在一条对角线上，则这个格子就在这个象的攻击范围内。 分析：简单分析可以得出：位于同一条主对角线上元素Map[i][j]满足i-j相同，位于同一条副对角线上的元素Map[i][j]满足i+j相同。如果两个象不会互相攻击，且攻击范围没有重合，那么这两个象的坐标x1+y1和x2+y2的奇偶性一定不同。所以我们只需预处理出每条对角线上的元素之和，然后更新即可。具体见代码。
#include#include 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; }