noip模拟赛 蒜头君打地鼠

分析：直接一个一个地去暴力枚举分数比较少，我们需要一种比较快的统计一定空间内1的数量，标准做法是前缀和，但是二维前缀和维护的是一个矩形内的值，这个是旋转过的该怎么办?可以把图旋转45°，不过这样比较考验码力，我们可以考虑维护每一行的前缀和，写得好常数小一点加上读入优化就能A了.

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

using namespace std;

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

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

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

    return ;
}