熱身訓練-k皇後問題(主副對角線計算)
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
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?≤?x
If there are several optimal solutions, you can print any of them.
Example
Input4Output
1 1 1 1
2 1 1 0
1 1 1 0
1 0 0 1
12
2 2 3 2
此題思路就是計算每一個位置所能收獲的價值,用一個變量來存取最大值。利用的重點每一條主對角線的x-y是一個定值而每一個副對角線之和x+y是一個定值;
#include<stdio.h>
const int N = 2005;
typedef long long LL;
LL Map[N][N];
LL L[N*2];//主對角線
LL R[N*2];//副對角線
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];//主對角線下x,y之和為定值n為確保不會出現負值
R[i+j] += Map[i][j];//副對角線x,y之差為定值
}
}
//顯然比下文常規算法更簡短易懂
/*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);
}
熱身訓練-k皇後問題(主副對角線計算)