分析：直接一個一個地去暴力枚舉分數比較少，我們需要一種比較快的統計一定空間內1的數量，標準做法是前綴和，但是二維前綴和維護的是一個矩形內的值，這個是旋轉過的該怎麽辦?可以把圖旋轉45°，不過這樣比較考驗碼力，我們可以考慮維護每一行的前綴和，寫得好常數小一點加上讀入優化就能A了.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

long long ans = 0,n, m, sum[2010][2010
 
];

long long read()
{
    long long res = 0, f = 1;
    char ch = getchar();
    while (ch < ‘0‘ || ch > ‘9‘)
        if (ch == ‘-‘)
        {
            f = -1;
            ch = getchar();
        }
    while (ch >= ‘0‘ && ch <= ‘9‘)
    {
        res = res * 10 + ch - ‘0‘;
        ch 
 
= getchar();
    }
    return res * f;
}

int main()
{
    n = read();
    m = read();
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
        {
            sum[i][j] = read();
            sum[i][j] += sum[i][j - 1];
        }
    for (int i = 1; i <= n; i++)
        for
 
 (int j = 1; j <= n; j++)
        {
            long long maxx = sum[i][min(n, j + m - 1)] - sum[i][max((long long)0, j - m)];
            for (int k = 1; k < m; k++)
            {
                int l = max(j - m + k,(long long)0), r = min(j + m - 1 - k,n);
                if (i + k <= n)
                    maxx += sum[i + k][r] - sum[i + k][l];
                if (i - k >= 1)
                    maxx += sum[i - k][r] - sum[i - k][l];
            }
            ans = max(maxx, ans);
        }
    printf("%lld\n", ans); 

    return 0;
}

noip模擬賽 蒜頭君打地鼠