洛谷P5300 与或和(全1子矩阵/单调栈）

洛谷P5300 与或和

题目链接

按照二进制将矩阵转换为32个01矩阵，and和就是一个01矩阵中全1子矩阵的个数乘以当前矩阵的贡献值，or和就是(总子矩阵个数-全0子矩阵的个数)*当前矩阵的贡献值。

#include "bits/stdc++.h"

using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const int maxn = 1e3 + 100;
bool d[35][maxn][maxn];
ll num[maxn][maxn];
ll st[maxn], tot, up[maxn], down[maxn];
int n;

void getnum(int k, int op) {//op为0时计算1的个数
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            num[i][j] = (d[k][i][j] ^ op) ? num[i][j - 1] + 1 : 0;
        }
    }
}

ll getcnt() {
    ll ret = 0;
    for (int j = 1; j <= n; j++) {
        tot = 0;
        for (int i = 1; i <= n; i++) {
            if (num[i][j]) {
                up[i] = 1;
                while (tot && num[i][j] <= num[st[tot]][j]) {
                    up[i] += up[st[tot]];
                    tot--;
                }
                st[++tot] = i;
            } else {
                up[i] = 0;
                tot = 0;
            }
        }
        tot = 0;
        for (int i = n; i >= 1; i--) {
            if (num[i][j]) {
                down[i] = 1;
                while (tot && num[i][j] < num[st[tot]][j]) {
                    down[i] += down[st[tot]];
                    tot--;
                }
                st[++tot] = i;
            } else {
                down[i] = 0;
                tot = 0;
            }
            ret = (ret + up[i] * down[i] * num[i][j] % mod) % mod;
        }
    }
    return (ret + mod) % mod;
}

ll p[35];

int main() {
    //freopen("in.txt", "r", stdin);
    cin >> n;
    ll x, nn = 0;
    p[0] = 1;
    for (int i = 1; i <= 32; i++) {
        p[i] = p[i - 1] * 2 % mod;
    }

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            nn = (nn + i * j) % mod;
            cin >> x;
            for (int k = 0; k <= 32; k++) {
                if (x & (1ll << k)) {
                    d[k][i][j] = true;
                }
            }
        }
    }
    ll ansand = 0, ansor = 0;
    for (int k = 0; k <= 32; k++) {
        getnum(k, 0);
        ansand = (ansand + p[k] * getcnt() % mod) % mod;
        getnum(k, 1);
        ansor = (ansor + p[k] * (nn - getcnt() + mod) % mod) % mod;
    }
    cout << ansand << " " << ansor << endl;
    return 0;
}