feat: 实现炮兵阵地问题的动态规划解法

添加动态规划解决方案来计算炮兵阵地的最大部署数量。使用状态压缩和预处理来优化性能，处理输入地图并计算合法状态及其对应的炮兵数量。

src/8/31/P2704.cpp

@@ -1,3 +1,81 @@
-int main(){
-    
-}
+#include <algorithm>
+#include <iostream>
+#include <string>
+#include <vector>
+
+
+static inline constexpr int popcount(int x) {
+    return __builtin_popcount(x);
+}
+
+int main() {
+    std::ios_base::sync_with_stdio(false);
+    std::cin.tie(NULL);
+
+    int n, m;
+    std::cin >> n >> m;
+
+    std::vector<int> map(n + 1, 0);
+    for (int i = 1; i <= n; ++i) {
+        std::string row;
+        std::cin >> row;
+        for (int j = 0; j < m; ++j) {
+            if (row[j] == 'H') {
+                map[i] |= (1 << j);
+            }
+        }
+    }
+
+    std::vector<int> states;
+    std::vector<int> counts;
+    for (int i = 0; i < (1 << m); ++i) {
+
+        if (!((i << 1) & i) && !((i << 2) & i)) {
+            states.push_back(i);
+            counts.push_back(popcount(i));
+        }
+    }
+    int num = states.size();
+
+    std::vector<std::vector<std::vector<int>>> dp(n + 1, std::vector<std::vector<int>>(num, std::vector<int>(num, 0)));
+
+    for (int i = 1; i <= n; ++i) {
+
+        for (int j = 0; j < num; ++j) {
+            int si = states[j];
+
+            if (si & map[i]) {
+                continue;
+            }
+
+            for (int k = 0; k < num; ++k) {
+                int sim1 = states[k];
+
+                if (si & sim1) {
+                    continue;
+                }
+
+                for (int l = 0; l < num; ++l) {
+                    int sim2 = states[l];
+
+                    if ((si & sim2) || (sim1 & sim2)) {
+                        continue;
+                    }
+
+                    dp[i][j][k] = std::max(dp[i][j][k], dp[i - 1][k][l] + counts[j]);
+                }
+            }
+        }
+    }
+
+    int ans = 0;
+    for (int j = 0; j < num; ++j) {
+        for (int k = 0; k < num; ++k) {
+            ans = std::max(ans, dp[n][j][k]);
+        }
+    }
+
+    std::cout << ans << std::endl;
+
+    return 0;
+}