HDU 6514 前缀和

题解

传送门 HDU 6514 Monitor

题解

对每个监视器所覆盖的矩形区域进行二维差分，取前缀和所得到的 $su{m}_{i,j}$ 代表点 $(i,j)$ 被覆盖的监视器数量，因为只考虑点是否被覆盖，那么取 $su{m}_{i,j}=sign(su{m}_{i,j})$。为了 $O(1)$ 求解偷盗者覆盖矩形区域是否都覆盖了监视器，对 $su{m}_{i,j}$ 取前缀和得到 $su{m}_{i,j}^{\prime }$，代表 $[1,i]\times [1,j]$ 区内域覆盖了监视器的点数。判断区域内覆盖了监视器的点数是否等于区域内总点数即可。总时间复杂度 $O(NM)$。

#include <bits/stdc++.h>
using namespace std;
typedef vector<int> vec;
typedef vector<vec> mat;
int N, M, P, Q;

int main()
{
    
    
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    while (cin >> N >> M)
    {
    
    
        mat a(N + 2, vec(M + 2));
        cin >> P;
        for (int i = 0; i < P; ++i)
        {
    
    
            int x1, y1, x2, y2;
            cin >> x1 >> y1 >> x2 >> y2;
            ++a[x1][y1], --a[x1][y2 + 1], --a[x2 + 1][y1], ++a[x2 + 1][y2 + 1];
        }
        for (int i = 1; i <= N; ++i)
            for (int j = 1; j <= M; ++j)
                a[i][j] += a[i - 1][j] + a[i][j - 1] - a[i - 1][j - 1];
        for (int i = 1; i <= N; ++i)
            for (int j = 1; j <= M; ++j)
                a[i][j] = (a[i][j] > 0) + a[i - 1][j] + a[i][j - 1] - a[i - 1][j - 1];
        cin >> Q;
        for (int i = 0; i < Q; ++i)
        {
    
    
            int x1, y1, x2, y2;
            cin >> x1 >> y1 >> x2 >> y2;
            int tar = (y2 - y1 + 1) * (x2 - x1 + 1);
            int sum = a[x2][y2] - a[x2][y1 - 1] - a[x1 - 1][y2] + a[x1 - 1][y1 - 1];
            cout << (sum == tar ? "YES" : "NO") << '\n';
        }
    }
    return 0;
}