实用指南：GESP2025年9月认证C++四级( 第三部分编程题（1）排兵布阵）

参考程序1（完全暴力）

#include 
using namespace std;
const int N = 15; // n,m <= 12，留一点余量
int a[N][N];
int main() {int n, m;cin >> n >> m;    //录入数据for (int i = 1; i <= n; ++i)for (int j = 1; j <= m; ++j)cin >> a[i][j];int ans = 0;// 枚举矩形：上(u)、左(l)、下(d)、右(r)for (int u = 1; u <= n; ++u) {for (int l = 1; l <= m; ++l) {for (int d = u; d <= n; ++d) {for (int r = l; r <= m; ++r) {bool ok = true;// 检查矩形 [u..d] x [l..r] 是否全为1for (int i = u; i <= d && ok; ++i) {for (int j = l; j <= r; ++j) {if (a[i][j] == 0) { ok = false; break; }}}if (ok) {int area = (d - u + 1) * (r - l + 1);if (area > ans) ans = area;}}}}}cout << ans << "\n";return 0;
}

参考程序2（优化暴力）

#include 
using namespace std;
const int N = 15;// n,m <= 12，留一点余量
int n, m;
int a[N][N];
int ans;
int main() {scanf("%d%d", &n, &m);for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)scanf("%d", &a[i][j]);for (int u = 1; u <= n; u++)for (int l = 1; l <= m; l++)for (int d = u; d <= n; d++) {int chk = 1;for (int r = l; r <= m; r++) {for (int x = u; x <= d; x++)chk &= a[x][r];if (!chk) break;ans = max(ans, (r - l + 1) * (d - u + 1));}}printf("%d\n", ans);return 0;
}

参考程序3（直方图+单调栈）

#include 
using namespace std;
int main() {int n, m;cin >> n >> m;vector> a(n, vector(m));for (int i = 0; i < n; ++i)for (int j = 0; j < m; ++j)cin >> a[i][j];vector h(m, 0);int ans = 0;for (int i = 0; i < n; ++i) {// 更新高度数组for (int j = 0; j < m; ++j) {if (a[i][j] == 1) h[j] += 1;else h[j] = 0;}// 求当前直方图最大矩形（单调栈，经典做法）stack st;for (int j = 0; j <= m; ++j) {// 当 j==m 时，视为高度 0，促使清栈int curH = (j == m ? 0 : h[j]);while (!st.empty() && h[st.top()] > curH) {int height = h[st.top()]; st.pop();int left = st.empty() ? -1 : st.top();int width = j - left - 1;ans = max(ans, height * width);}st.push(j);}}cout << ans << "\n";return 0;
}