feat: 添加P8903题解实现动态规划解法

实现了一个动态规划解决方案来计算最大收益，处理两种不同成本的物品选择。通过预处理前后缀最大值来优化计算效率。

src/10/9/P8903.cpp

@@ -0,0 +1,87 @@
+#include <algorithm>
+#include <cstdint>
+#include <iostream>
+#include <istream>
+using ll = int64_t;
+
+const ll maxn = 2005;
+
+struct C {
+    ll p, c, x;
+    bool operator<(const C &other) const {
+        return x < other.x;
+    }
+} c[maxn];
+
+ll n, a, b;
+ll dpb[maxn][maxn], dpa[maxn][maxn];
+
+ll mxb[maxn][maxn], mxa[maxn][maxn];
+
+int main() {
+    std::iostream::sync_with_stdio(false);
+    std::cin.tie(nullptr);
+
+    std::cin >> n >> a >> b;
+    for (ll i = 1; i <= n; i++) {
+        std::cin >> c[i].p >> c[i].c >> c[i].x;
+    }
+    std::sort(c + 1, c + 1 + n);
+
+    for (ll i = 1; i <= n; i++) {
+        for (ll j = 0; j <= b; j++) {
+            dpb[i][j] = dpb[i - 1][j];
+            if (j >= c[i].c * c[i].x) {
+                dpb[i][j] = std::max(dpb[i][j], dpb[i - 1][j - c[i].c * c[i].x] + c[i].p);
+            }
+        }
+    }
+
+    for (ll i = 1; i <= n; i++) {
+        for (ll j = 0; j <= b; j++) {
+            if (j > 0) {
+                mxb[i][j] = std::max(mxb[i][j - 1], dpb[i][j]);
+            } else {
+                mxb[i][j] = dpb[i][j];
+            }
+        }
+    }
+
+    for (ll i = n; i >= 1; i--) {
+        for (ll j = 0; j <= a; j++) {
+            dpa[i][j] = dpa[i + 1][j];
+            if (j >= c[i].c) {
+                dpa[i][j] = std::max(dpa[i][j], dpa[i + 1][j - c[i].c] + c[i].p);
+            }
+        }
+    }
+
+    for (ll i = n; i >= 1; i--) {
+        for (ll j = 0; j <= a; j++) {
+            if (j > 0) {
+                mxa[i][j] = std::max(mxa[i][j - 1], dpa[i][j]);
+            } else {
+                mxa[i][j] = dpa[i][j];
+            }
+        }
+    }
+
+    ll ans = 0;
+
+    ans = std::max(ans, mxb[n][b]);
+    ans = std::max(ans, mxa[1][a]);
+
+    for (ll i = 1; i <= n; i++) {
+        for (ll j = 0; j <= c[i].c; j++) {
+            ll mcost = c[i].c - j;
+            ll iccost = j * c[i].x;
+            if (mcost <= a && iccost <= b) {
+
+                ll pre = (i > 1) ? mxb[i - 1][b - iccost] : 0;
+                ll suff = (i < n) ? mxa[i + 1][a - mcost] : 0;
+                ans = std::max(ans, pre + c[i].p + suff);
+            }
+        }
+    }
+    std::cout << ans << "\n";
+}