【CodeForces】Gargari and Bishops

依据贪心能够知道，放置的教主必须不能相互攻击到（也就是不在一条对角线上）才干够使得结果最大化。

依据观察能够得到教主相互不攻击的条件是他的坐标和互为奇偶（x + y)

之后直接暴力，处理每一个坐标对角线的和就好

时间复杂度 0(n ^ 2)

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn = 2005;
LL sum_l[maxn * 10] = {0},sum_r[maxn * 10] = {0};
LL mat[maxn][maxn];
int main(){
    int n,m;
    LL  x;
    scanf("%d",&n);
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++){
            scanf("%I64d",&mat[i][j]);
            sum_l[i + j] += mat[i][j];
            sum_r[n - 1 + j - i] += mat[i][j];
        }
    }
    //for(int i = 0; i < 2 * n - 1; i++)
    //    printf("%I64d %I64d\n",sum_l[i],sum_r[i]);
    LL max1 = - 1,max2 = -1;
    int x1,x2,y1,y2;
    for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++){
            LL ret = sum_l[i + j] + sum_r[n - 1 + j - i] - mat[i][j];
            if(((i + j) & 1) && ret > max1){
                max1 = ret;
                x1 = i;
                y1 = j;
            }
            else if((!((i + j) & 1)) && ret > max2){
                max2 = ret;
                x2 = i;
                y2 = j;
            }
        }
    LL ans = max1 + max2;
    printf("%I64d\n",ans);
    printf("%d %d %d %d\n",x1 + 1,y1 + 1,x2 + 1,y2 + 1);
    return 0;
}