P2213 [USACO14MAR]懒惰的牛The Lazy Cow_Sliver

利用之前算过的信息，每个点线性转移上个点的值，扣掉两个斜行，加上两个斜行即可。 专门写一篇题解是因为之前有一次考试写挂了类似的题。

#include <cstdio>
#include <cstring>
#include <cassert>
#include <iostream>
#include <algorithm>
using namespace std;
const int MAXN = 4e2 + 10;
inline int read()
{
    int x = 0; char ch = getchar(); bool f = false;
    while(!isdigit(ch)) f |= (ch == '-'), ch = getchar();
    while(isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f ? -x : x;
}

int N, K;
int a[MAXN][MAXN], sum[MAXN][MAXN];

inline bool out(int x, int y){
    return !(1 <= x && x <= N && 1 <= y && y <= N);
}

int main()
{
    // freopen("p2213.in", "r", stdin);
    cin>>N>>K;
    for(int i = 1; i <= N; i++)
        for(int j = 1; j <= N; j++)
            a[i][j] = read();
    for(int i = 1; i <= N; i++)
         for(int j = 1; j <= N; j++)
            sum[i][j] = sum[i - 1][j] + a[i][j];

    int ans = 0;
    for(int i = 1; i <= N; i++){
        int res = 0;
        for(int j = 0; j <= min(K, N - 1); j++)
            res += sum[min(i + K - j, N)][1 + j] - sum[max(i - K + j, 1) - 1][1 + j];
        ans = max(ans, res);
        // cout<<res<<" ";
        for(int j = 2; j <= N; j++){
            for(int k = K; k >= 0; k--){
                int x = i + k, y = j + K - k;
                if(!out(x, y)) res += a[x][y];
                x = i - k;
                if(!out(x, y)) res += a[x][y];
            }
            if(!out(i, j + K)) res -= a[i][j + K];
            for(int k = K; k >= 0; k--){
                int x = i - k, y = j - 1 - K + k;
                if(!out(x, y)) res -= a[x][y];
                x = i + k;
                if(!out(x, y)) res -= a[x][y];
            }
            if(!out(i, j - K - 1)) res += a[i][j - K - 1];
            // cout<<res<<" ";
            ans = max(ans, res);
        }
        // cout<<endl;
    }
    cout<<ans<<endl;
    return 0;
}