一个能运行但是会T的版本，因为本质上还是\(O(nmab)\)的算法。每次\(O(ab)\)初始化矩阵中的可能有用的点，然后\(O(n-a)\)往下推。

#include
    
     using namespace std;typedef long long ll;#define ERR(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator
     
       _it(_ss); err(_it, args); }void err(istream_iterator
      
        it) {cerr << "\n";}template
       
        void err(istream_iterator
        
          it, T a, Args... args) {    cerr << *it << "=" << a << ", ";    err(++it, args...);}#define ERR1(arg,n) { cerr<<""<<#arg<<"=\n  "; for(int i=1;i<=n;i++) cerr<
         
          <<" "; cerr<<"\n"; }#define ERR2(arg,n,m) { cerr<<""<<#arg<<"=\n"; for(int i=1;i<=n;i++) { cerr<<" "; for(int j=1;j<=m;j++)cerr<
          
           <<" "; cerr<<"\n"; } }int n, m, a, b;int x, y, z;int g[3000 * 3000 + 5];int h[3005][3005];struct node { int v, x; node(int vv, int xx) { v = vv; x = xx; }};int curx, cury;deque
           
             dq;void calc(int x, int y) { curx = x, cury = y; while(!dq.empty()) dq.pop_back(); for(int i = 1; i <= a; i++) { int minline = h[x + i - 1][y]; for(int j = 2; j <= b; j++) { minline = min(minline, h[x + i - 1][y + j - 1]); } while(!dq.empty() && minline <= dq.back().v) { dq.pop_back(); } if(dq.empty() || minline > dq.back().v) { dq.push_back(node(minline, x + i - 1)); } }}void move_to_nextline() { curx++; if(dq.front().x < curx) dq.pop_front(); int minline = h[curx + a - 1][cury]; for(int j = 2; j <= b; j++) { minline = min(minline, h[curx + a - 1][cury + j - 1]); } while(!dq.empty() && minline <= dq.back().v) { dq.pop_back(); } if(dq.empty() || minline > dq.back().v) { dq.push_back(node(minline, curx + a - 1)); }}ll ans = 0;int main() {#ifdef Yinku freopen("Yinku.in", "r", stdin); //freopen("Yinku.out", "w", stdout);#endif // Yinku while(~scanf("%d%d%d%d", &n, &m, &a, &b)) { scanf("%d%d%d%d", &g[1], &x, &y, &z); for(int i = 2; i <= n * m; i++) g[i] = (1ll * g[i - 1] * x % z + y) % z; for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) h[i][j] = g[(i - 1) * m + j]; //ERR2(h, n, m); ans = 0; for(int i = 1; i + a - 1 <= n; i++) { for(int j = 1; j + b - 1 <= m; j++) { calc(i, j); ans += dq.front().v; for(int di = 1; di <= a; di++) { move_to_nextline(); } //ERR(ans); } } printf("%lld\n", ans); }}
           
          
         
        
       
      
     
    

其实不需要重复计算这么多的单调队列。

具体的思路是这样：

先把左侧的n行b列插入各行的单调队列dq[i]，然后取出各个队列的队首竖着组成单调队列dq2，这个单调队列dq2就可以回答左侧n行b列的所有的最小值，复杂度O(nb)。

向右移动n个dq[i]，再回答左侧n行，[2,b+1]列的所有的最小值，复杂度O(n)。

总体复杂度O(nm)。

先用STL写了一个

#include
    
     using namespace std;typedef long long ll;namespace Debug {#define ERR(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator
     
       _it(_ss); err(_it, args); }    void err(istream_iterator
      
        it) {cerr << "\n";}    template
       
            void err(istream_iterator
        
          it, T a, Args... args) {        cerr << *it << "=" << a << ", ";        err(++it, args...);    }#define ERR1(arg,n) { cerr<<""<<#arg<<"=\n  "; for(int i=1;i<=n;i++) cerr<
         
          <<" "; cerr<<"\n"; }#define ERR2(arg,n,m) { cerr<<""<<#arg<<"=\n"; for(int i=1;i<=n;i++) { cerr<<" "; for(int j=1;j<=m;j++)cerr<
          
           <<" "; cerr<<"\n"; } }}int n, m, a, b;int x, y, z;int g[3005 * 3005];int h[3005][3005];ll ans;struct Node { int val; int id; Node() {} Node(int val, int id): val(val), id(id) {} friend bool operator>=(const Node& n, const int& v) { return n.val >= v; } friend bool operator>=(const Node& n1, const Node& n2) { return n1.val >= n2.val; }};deque
           
             dq[3005];void init_deque(int i) { dq[i].clear(); for(int j = 1; j <= b; j++) { while(!dq[i].empty() && dq[i].back() >= h[i][j]) { dq[i].pop_back(); } dq[i].push_back({h[i][j], j}); }}void move_deque(int j) { for(int i = 1; i <= n; i++) { if(dq[i].front().id < j - b + 1) { dq[i].pop_front(); } while(!dq[i].empty() && dq[i].back() >= h[i][j]) { dq[i].pop_back(); } dq[i].push_back({h[i][j], j}); }}deque
            
              dq2;void calc_ans(int oj) { dq2.clear(); for(int i = 1; i <= a; i++) { while(!dq2.empty() && dq2.back() >= dq[i].front()) dq2.pop_back(); dq2.push_back({dq[i].front().val, i}); } ans += dq2.front().val; for(int i = a + 1; i <= n; i++) { if(dq2.front().id < i - a + 1) dq2.pop_front(); while(!dq2.empty() && dq2.back() >= dq[i].front()) dq2.pop_back(); dq2.push_back({dq[i].front().val, i}); ans += dq2.front().val; }}int main() {#ifdef Yinku freopen("Yinku.in", "r", stdin); //freopen("Yinku.out", "w", stdout);#endif // Yinku while(~scanf("%d%d%d%d", &n, &m, &a, &b)) { scanf("%d%d%d%d", &g[1], &x, &y, &z); for(int i = 2; i <= n * m; i++) g[i] = (1ll * g[i - 1] * x % z + y) % z; for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) h[i][j] = g[(i - 1) * m + j]; for(int i = 1; i <= n; i++) { init_deque(i); } ans = 0; calc_ans(b); for(int nj = b + 1; nj <= m; nj++) { move_deque(nj); calc_ans(nj); } printf("%lld\n", ans); }}