feat: 实现石子合并问题的动态规划解法

添加完整的动态规划实现来计算石子合并问题的最小和最大得分
优化输入输出性能并修复原未实现的代码

src/8/27/P1880.cpp

@@ -1,32 +0,0 @@
-/*
-
-
-
-*/
-#include <cstdint>
-#include <iostream>
-#include <stdexcept>
-#include <vector>
-using ll = int64_t;
-
-int main(){
-    ll n;
-    std::cin>>n;
-    std::vector<ll> arr(n*2+1);
-    std::vector<ll> pre(n*2+1);
-    for(ll i=1;i<=n;i++){
-        std::cin>>arr[i];
-        arr[n+i]=arr[i];
-    }
-    for(ll i=1;i<=2*n;i++){
-        pre[i]=pre[i-1]+arr[i];
-    }
-    std::vector<std::vector<ll>> dp;
-    for(ll i=1;i<n;i++){
-        ll j=i+n-1;
-        dp.clear();
-        dp.resize(2*n+1,std::vector<ll>(2*n+1,1e9));
-        // for(ll x=1;)
-        throw std::runtime_error("NotImpletmented");
-    }
-}

src/8/29/P1880.cpp

@@ -0,0 +1,54 @@
+#include <algorithm>
+#include <cstdint>
+#include <iostream>
+#include <istream>
+#include <vector>
+using ll = int64_t;
+
+const ll inf=1e9;
+
+int main(){
+    std::iostream::sync_with_stdio(false);
+    std::cin.tie(nullptr);
+    std::cout.tie(nullptr);
+
+    ll n;
+    std::cin>>n;
+    std::vector<std::vector<ll>> dpmax(2*n+1,std::vector<ll>(2*n+1,-inf)),
+                                    dpmin(2*n+1,std::vector<ll>(2*n+1,inf)),
+                                    cost(2*n+1,std::vector<ll>(2*n+1));
+
+    for(ll i=1;i<=n;i++){
+        std::cin>>cost[i][i];
+        cost[n+i][n+i]=cost[i][i];
+        dpmin[i][i]=dpmax[i][i]=0;
+        dpmin[n+i][n+i]=dpmax[n+i][n+i]=0;
+    }
+
+    for(ll len=2;len<=n;++len){
+        for(ll i=1;i<=2*n-len;i++){
+            ll j=i+len-1;
+            for(ll k=i;k<j;k++){
+                cost[i][j]=cost[i][k]+cost[k+1][j];
+            }
+        }
+    }
+
+    for(ll len=2;len<=n;len++){
+        for(ll i=1;i<=2*n-len;i++){
+            ll j=i+len-1;
+            for(ll k=i;k<j;k++){
+                dpmax[i][j]=std::max(dpmax[i][j],dpmax[i][k]+dpmax[k+1][j]+cost[i][j]);
+                dpmin[i][j]=std::min(dpmin[i][j],dpmin[i][k]+dpmin[k+1][j]+cost[i][j]);
+            }
+        }
+    }
+
+    ll minans=inf,maxans=-inf;
+    for(ll i=1;i<=n;i++){
+        minans=std::min(minans,dpmin[i][i+n-1]);
+        maxans=std::max(maxans,dpmax[i][i+n-1]);
+    }
+
+    std::cout<<minans<<"\n"<<maxans<<"\n";
+}