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 <bits/stdc++.h>
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;
}

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 <bits/stdc++.h>
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;
}