思路：

预处理好前缀和，枚举上边界和下边界，将二维变成一维，用单调队列找满足题意的最小前缀

复杂度，O（r*r*c）

代码：

#pragma GCC optimize(2)#pragma GCC optimize(3)#pragma GCC optimize(4)#include
     
      using namespace std;#define fi first#define se second#define pi acos(-1.0)#define LL long long//#define mp make_pair#define pb push_back#define ls rt<<1, l, m#define rs rt<<1|1, m+1, r#define ULL unsigned LL#define pll pair
      
       #define pii pair
       
        #define piii pair
        
         #define mem(a, b) memset(a, b, sizeof(a))#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);//headconst int N = 555;const LL INF = 0x7f7f7f7f7f7f7f7f;int a[N][N];LL sum[N][N], tot[N];int sm[N][N], cnt[N];deque
         
          q;int main() {    int n, m, x, y, z;    scanf("%d %d %d %d %d", &n, &m, &x, &y, &z);    for (int i = 1; i <= n; i++) {        for (int j = 1; j <= m; j++)            scanf("%d", &a[i][j]);    }    for (int j = 1; j <= m; j++) {        for (int i = 1; i <= n; i++) {            sum[i][j] = sum[i-1][j] + a[i][j];            sm[i][j] = sm[i-1][j] + (a[i][j] == 0);        }    }    LL ans = -INF;    for (int l = 1; l <= n; l++) {        for (int r = l; r <= n && r <= l+x-1; r++) {            q.clear();            tot[0] = 0;            cnt[0] = 0;            q.push_back(0);            for (int i = 1; i <= m; i++) {                tot[i] = tot[i-1] + sum[r][i] - sum[l-1][i];                cnt[i] = cnt[i-1] + sm[r][i] - sm[l-1][i];                while(!q.empty() && tot[q.back()] >= tot[i]) q.pop_back();                q.push_back(i);                while(!q.empty() && q.front() < i - y) q.pop_front();                while(!q.empty() && cnt[i] - cnt[q.front()] > z) q.pop_front();                if(!q.empty()) ans = max(ans, tot[i] - tot[q.front()]);            }        }    }    printf("%lld\n", ans);    return 0;}