Wannafly挑战赛19 B矩阵

矩阵

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<bits/stdc++.h>
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<LL, LL>
#define pii pair<int, int>
#define piii pair<pii, int>
#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);
//head
 
const int N = ;
const LL INF = 0x7f7f7f7f7f7f7f7f;
int a[N][N];
LL sum[N][N], tot[N];
int sm[N][N], cnt[N];
deque<int>q;
int main() {
    int n, m, x, y, z;
    scanf("%d %d %d %d %d", &n, &m, &x, &y, &z);
    for (int i = ; i <= n; i++) {
        for (int j = ; j <= m; j++)
            scanf("%d", &a[i][j]);
    }
 
    for (int j = ; j <= m; j++) {
        for (int i = ; i <= n; i++) {
            sum[i][j] = sum[i-][j] + a[i][j];
            sm[i][j] = sm[i-][j] + (a[i][j] == );
        }
    }
 
    LL ans = -INF;
    for (int l = ; l <= n; l++) {
        for (int r = l; r <= n && r <= l+x-; r++) {
            q.clear();
            tot[] = ;
            cnt[] = ;
            q.push_back();
            for (int i = ; i <= m; i++) {
                tot[i] = tot[i-] + sum[r][i] - sum[l-][i];
                cnt[i] = cnt[i-] + sm[r][i] - sm[l-][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 ;
}