Codeforces 1195E OpenStreetMap

题目链接：https://codeforces.com/contest/1195/problem/E

考虑先对每行做一次单调队列，把长度为$a$所有最小值存起来，那么对于得到的新数组，我们只需要将它在列方向上对每列做一次单调队列，那么得到的即是答案。

#include <bits/stdc++.h>
#define pii pair<ll, ll>
#define pil pair<ll, long long>
#define pll pair<long long, long long>
#define lowbit(x) ((x)&(-x))
#define mem(i, a) memset(i, a, sizeof(i))
#define sqr(x) ((x)*(x))
#define all(x) x.begin(),x.end()
#define ls (k << 1)
#define rs (k << 1 | 1)
using namespace std;
typedef long long ll;
template <typename T>
inline void read(T &X) {
    X = 0; char ch = 0; T op = 1;
    for(; ch > '9' || ch < '0'; ch = getchar())
        if(ch == '-') op = -1;
    for(; ch >= '0' && ch <= '9'; ch = getchar())
        X = (X << 3) + (X << 1) + ch - 48;
    X *= op;
}

const ll INF = 0x3f3f3f3f;
const ll N = 3005 + 5;
ll s[N][N],h[N][N];
ll g[N * N];
ll q[N];
int main() {
#ifdef INCTRY
    freopen("input.txt", "rt", stdin);
#endif
    ll n,m,a,b;
    cin >> n >> m >> a >> b;
    ll x,y,z;
    cin >> g[0] >> x >> y >> z;
    for(ll i = 1; i <= n * m; i++) {
        g[i] = g[i - 1] * x + y;
        g[i] %= z;
    }
    for(ll i = 1; i <= n; i++)
        for(ll j = 1; j <= m; j++)
            h[i][j] = g[(i-1)*m+j-1];
    ll ans = 0;
    for(ll i = 1; i <= n; i++) {
        ll l = 1, r = 0;
        for(ll j = 1; j <= m; j++) {
            while(l <= r && q[l] <= j - b) l++;
            while(l <= r && h[i][j] <= h[i][q[r]]) r--;
            q[++r] = j;
            if(j >= b) s[i][j - b + 1] = h[i][ q[l] ];
        }
    }

    for(ll j = 1; j <= m - b + 1; j++) {
        ll l = 1, r = 0;
        for(ll i = 1; i <= n; i++) {
            while(l <= r && q[l] <= i - a) l++;
            while(l <= r && s[i][j] <= s[q[r]][j]) r--;
            q[++r] = i;
            if(i >= a) ans += s[q[l]][j];
        }
    }
    cout << ans;

#ifdef INCTRY
    cerr << "\nTime elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
    return 0;
}