Codeforces 263E Rhombus （看题解)

Rhombus

不想写标程啊， 转成切比雪夫距离暴力就能过啦。。 复杂度n * m * k， 其实复杂度能处理到n * m

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 2000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

int n, m, k;
int a[N][N];
LL b[N][N];

inline LL getVal(int X1, int X2, int Y1, int Y2) {
    return b[X2][Y2] - b[X2][Y1 - 1] - b[X1 - 1][Y2] + b[X1 - 1][Y1 - 1];
}

int main() {
    scanf("%d%d%d", &n, &m, &k);
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= m; j++) {
            scanf("%d", &a[i][j]);
            b[i + j][i - j + m] = a[i][j];
        }
    }
    for(int i = 1; i <= n + m; i++)
        for(int j = 1; j <= n + m; j++)
            b[i][j] += b[i - 1][j] + b[i][j - 1] - b[i - 1][j - 1];
    LL ans = -INF, ret = 0;
    int ansx, ansy, x, y;
    for(int i = k; i <= n - k + 1; i++) {
        for(int j = k; j <= m - k + 1; j++) {
            x = i + j, y = i - j + m;
            ret = 0;
            for(int z = 0; z < k; z++)
                ret += getVal(x - z, x + z, y - z, y + z);
            if(chkmax(ans, ret)) ansx = i, ansy = j;
        }
    }
    printf("%d %d\n", ansx, ansy);
    return 0;
}

/*
*/

原文地址：https://www.cnblogs.com/CJLHY/p/10905211.html