AtCoder ABC260 G 累积和

题意

传送门 AtCoder ABC260 G Scalene Triangle Area

题解

暴力 + 前缀和

对每一行计算前缀和，处理每一个询问时暴力枚举每一行进行累加。总时间复杂度 $O(n(n+q))$。

#include 
using namespace std;

void solve() {
    int n, m;
    cin >> n >> m;
    vector<string> ss(n);
    for (int i = 0; i < n; ++i) {
        cin >> ss[i];
    }
    vector<vector<int>> sum(n, vector<int>(n + 1));
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            sum[i][j + 1] = sum[i][j] + (ss[i][j] == 'O');
        }
    }
    int q;
    cin >> q;
    while (q--) {
        int u, v;
        cin >> u >> v;
        u -= 1;
        v -= 1;
        int res = 0;
        for (int s = u; s > max(-1, u - m); --s) {
            res += sum[s][v + 1] - sum[s][max(-1, v - 2 * (m + s - u)) + 1];
        }
        cout << res << '\n';
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    solve();

    return 0;
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39

累积和

根据 Tutorial 的累积和做法，可以不断根据某一维度做差分，直到差分需要处理的规模适当的小；甚至可以按照不同维度分别处理贡献后再累计求和。总时间复杂度 $O(n^2)$。

#include 
using namespace std;

void solve() {
    int n, m;
    cin >> n >> m;
    vector<string> ss(n);
    for (int i = 0; i < n; ++i) {
        cin >> ss[i];
    }
    vector<vector<int>> a(n, vector<int>(n));
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            if (ss[i][j] == 'O') {
                a[i][j] += 1;
                if (i + m < n) {
                    a[i + m][j] -= 1;
                }
            }
        }
    }
    for (int j = 0; j < n; ++j) {
        for (int i = 0; i + 1 < n; ++i) {
            a[i + 1][j] += a[i][j];
        }
    }
    vector<vector<int>> b(n, vector<int>(5 * n));
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            if (ss[i][j] == 'O') {
                b[i][j + 2 * m] -= 1;
                if (i + m < n) {
                    b[i + m][j] += 1;
                }
            }
        }
    }
    for (int i = 0; i + 1 < n; ++i) {
        for (int j = 0; j + 2 < 5 * n; ++j) {
            b[i + 1][j] += b[i][j + 2];
        }
    }
    vector<vector<int>> sum(n, vector<int>(n));
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            sum[i][j] = a[i][j] + b[i][j];
        }
    }
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j + 1 < n; ++j) {
            sum[i][j + 1] += sum[i][j];
        }
    }
    int q;
    cin >> q;
    while (q--) {
        int x, y;
        cin >> x >> y;
        x -= 1, y -= 1;
        cout << sum[x][y] << '\n';
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    solve();

    return 0;
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71