Description

輸入一個n*m的矩陣，矩陣的每一個元素都是一個整數，然後有q個詢問，每次詢問一個子矩陣的權值。矩陣的權值是這樣定義的，對於一個整數x，如果它在該矩陣中出現了p次，那麼它給該矩陣的權值就貢獻p2。

Solution

由於出現$p$次的元素的貢獻是$p^2$，我們可以看做每一對相同的元素可以產生$1$的貢獻。

我們定一個$S$，大概為$40$。 對於出現次數大於$S$的元素，由於這樣的元素個數不多，直接暴力搞。 對於小於$S$的元素，我們兩兩配對，由於要求在矩形範圍內，有$4$個限制，所以是個四維偏序問題，將第一維排序，剩下的用三維樹狀陣列即可。

Code

#include
 
<bits/stdc++.h>
  
using namespace std;
  
typedef long long LL;
typedef long double LD;
typedef pair<int, int> PII;
#define Rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define
 
 Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define X first
#define Y second
#define PB(a) push_back(a)
#define MP(a, b) make_pair(a, b)
#define SZ(a) ((int)(a).size())
#define
 
 ALL(a) a.begin(), a.end()
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia
#define sqr(x) ((x) * (x))
  
template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }
  
inline int read() {
    static int _, __; static char c_;
    for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
    for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}
//EOT
  
  
const int maxn = 233, maxm = 233 * 233, S = 35, sumc = maxm / S + 5, maxq = 201314;
  
int n, m, q, ls[maxm], lss, ans[maxq], ty[maxm], cnt[maxm], a[maxn][maxn];
  
struct Query {
    int x1, y1, x2, y2;
}qry[maxq];
  
namespace FuckYouAll {
    int lss, ls[maxm], sum[maxn][maxn];
    inline void Main() {
        For(k, 1, lss) {
            For(i, 1, n) For(j, 1, m)
                sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + (a[i][j] == ls[k]);
            For(p, 1, q) {
                ans[p] += sqr(sum[qry[p].x2][qry[p].y2] + sum[qry[p].x1 - 1][qry[p].y1 - 1]
                        - sum[qry[p].x1 - 1][qry[p].y2] - sum[qry[p].x2][qry[p].y1 - 1]);
            }
        }
    }
}
  
namespace LoveYouForever {
    int lss, ls[maxn * maxn];
  
    struct Point { int a, b, c, d, id; }p[S * maxm + maxq];
    int ps;
    inline bool cmp(Point A, Point B) {
        if (A.a != B.a) return A.a < B.a;
        if (A.b != B.b) return A.b < B.b;
        if (A.c != B.c) return A.c < B.c;
        if (A.d != B.d) return A.d < B.d;
        return A.id < B.id;
    }
  
    struct BIT {
        int c[maxn][maxn][maxn];
#define lb(x) ((x) & -(x))
        inline void update(int x, int y, int z) {
            for (int i = x; i <= m; i += lb(i))
                for (int j = y; j <= n; j += lb(j))
                    for (int k = z; k <= m; k += lb(k))
                        ++ c[i][j][k];
        }
        inline int query(int x, int y, int z) {
            int ans = 0;
            for (int i = x; i; i -= lb(i))
                for (int j = y; j; j -= lb(j))
                    for (int k = z; k; k -= lb(k))
                        ans += c[i][j][k];
            return ans;
        }
    }bit;
  
    vector<PII> vct[maxm];
  
    inline void Main() {
        For(i, 1, n) For(j, 1, m) if (ty[a[i][j]] == 2)
            vct[lower_bound(ls + 1, ls + 1 + lss, a[i][j]) - ls].push_back(MP(i, j));
        For(i, 1, lss) {
            Rep(j, SZ(vct[i])) Rep(k, SZ(vct[i])) {
                int x1 = vct[i][j].X, y1 = vct[i][j].Y, x2 = vct[i][k].X, y2 = vct[i][k].Y;
                if (x1 > x2) swap(x1, x2);
                if (y1 > y2) swap(y1, y2);
                p[++ ps] = (Point){n - x1 + 1, m - y1 + 1, x2, y2, 0};
            }
        }
        For(i, 1, q)
            p[++ ps] = (Point){n - qry[i].x1 + 1, m - qry[i].y1 + 1, qry[i].x2, qry[i].y2, i};
        sort(p + 1, p + 1 + ps, cmp);
        //For(i, 1, ps) cout << p[i].a << ' ' << p[i].b << ' ' << p[i].c << ' ' << p[i].d << ' ' << p[i].id<< endl;
        For(i, 1, ps)
            if (!p[i].id) bit.update(p[i].b, p[i].c, p[i].d);
            else ans[p[i].id] += bit.query(p[i].b, p[i].c, p[i].d);
    }
}
  
int main()
{
#ifdef hany01
    freopen("vegetable.in", "r", stdin);
    freopen("vegetable.out", "w", stdout);
#endif
  
    n = read(), m = read();
    For(i, 1, n) For(j, 1, m) ls[(i - 1) * m + j] = a[i][j] = read();
    sort(ls + 1, ls + 1 + n * m), lss = unique(ls + 1, ls + 1 + n * m) - ls - 1;
    For(i, 1, n) For(j, 1, m) ++ cnt[a[i][j] = lower_bound(ls + 1, ls + 1 + lss, a[i][j]) - ls];
    For(i, 1, lss)
        if (cnt[i] > S) FuckYouAll::ls[++ FuckYouAll:: lss] = i, ty[i] = 1;
        else LoveYouForever:: ls[++ LoveYouForever:: lss] = i, ty[i] = 2;
	q = read();
    For(i, 1, q) {
		qry[i].x1 = read(), qry[i].y1 = read(), qry[i].x2 = read(), qry[i].y2 = read();
		if (qry[i].x1 > qry[i].x2) swap(qry[i].x1, qry[i].x2);
		if (qry[i].y1 > qry[i].y2) swap(qry[i].y1, qry[i].y2);
	}
  
    FuckYouAll:: Main();
    LoveYouForever:: Main();
    For(i, 1, q) printf("%d\n", ans[i]);
  
    return 0;
}