傳送門：http://www.lydsy.com/JudgeOnline/problem.php?id=3671

【題解】

貪心從1...n*m取，開兩個5000*5000的數組就夠了，可以重復利用，坐標可以壓到一個int裏。

每次暴力標記不能訪問的，標到已經有標記的就不用標了因為後面的肯定前面已經標記過了。

均攤復雜度就對了。復雜度$O(nm)$。

這破題還卡PE..

# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>
 


using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 5e3 + 1;
const int mod = 1e9+7;

int mp[M][M];
int a[M * M], xc, A, B, C, D, n, m, T;
inline int gnext(int x) {
    return (1ll * A * x % D * x % D + 1ll * B * x % D + C) % D;
}

# define id(i, j) (((i)
 
-1) * m + (j))

int main() {
    int Q;
    cin >> xc >> A >> B >> C >> D;
    cin >> n >> m >> Q; T = n*m;
    for (int i=1; i<=T; ++i) a[i] = i;
    for (int i=1; i<=T; ++i) {
        xc = gnext(xc);
        swap(a[i], a[xc % i + 1]);
    }
    
 
while(Q--) {
        scanf("%d%d", &A, &B);
        swap(a[A], a[B]);
    }
    for (int i=1; i<=n; ++i)
        for (int j=1; j<=m; ++j) 
            mp[i][j] = a[id(i, j)];
    for (int i=1; i<=n; ++i)
        for (int j=1; j<=m; ++j)
            a[mp[i][j]] = id(i, j), mp[i][j] = 0;
    int ux, uy, vx, vy, x, y, ok = 1;
    ux = n, uy = m, vx = 1, vy = 1;
    for (int i=1; i<=T; ++i) {
        x = (a[i] - 1) / m + 1;
        y = a[i] - (x-1) * m;
        if(!mp[x][y]) {
            if(ok) printf("%d", i), ok = 0;
            else printf(" %d", i); 
            for (int a=x+1; a<=n; ++a)
                for (int b=y-1; b; --b) {
                    if(mp[a][b]) break; 
                    mp[a][b] = 1; 
                }
            for (int a=x-1; a; --a)
                for (int b=y+1; b<=m; ++b) {
                    if(mp[a][b]) break;
                    mp[a][b] = 1;
                }
        }
    }
    puts("");
    return 0;
}

bzoj3671 [Noi2014]隨機數生成器