标签：print keyword ons mon att any des sts oar
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.
Example
4
1 1 1 1
2 1 1 0
1 1 1 0
1 0 0 1
12
2 2 3 2
此题思路就是计算每一个位置所能收获的价值，用一个变量来存取最大值。利用的重点每一条主对角线的x-y是一个定值而每一个副对角线之和x+y是一个定值；
#include<stdio.h>
const int N = 2005;
typedef long long LL;
LL Map[N][N];
LL L[N*2];//主对角线
LL R[N*2];//副对角线
int main()
{
int n;
scanf("%d",&n);
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= n; j++) {
scanf("%I64d",&Map[i][j]);
L[i-j+n] += Map[i][j];//主对角线下x，y之和为定值n为确保不会出现负值
R[i+j] += Map[i][j];//副对角线x，y之差为定值
}
}
//显然比下文常规算法更简短易懂
/*for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
m=0;
if(i>=j)
{
for(int x=0;x<n-i+j;x++)
{
m=m+a[i-j+x][x];
}
}
else
{
for(int x=0;x<n-j+i;x++)
{
m=m+a[x][j-i+x];
}
}
if(i+j<=n)
{
for(int x=0;x<i+j+1;x++)
{
m=m+a[x][i+j-x];
}
}
else
{
for(int x=n-1;x>i+j-n;x--)
{
m=m+a[x][i+j-x];
}
}
m=m-a[i][j];
if(m>m1)
{
m1=m;
x1=i+1;
y1=j+1;
}
}
}
*/
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);
}
标签：print keyword ons mon att any des sts oar
原文地址：http://www.cnblogs.com/tigerzhou/p/7106599.html
