update

README.md

@@ -0,0 +1,10 @@
+## 线性动态规划优化为$O(n\log{n})$方法
+>如果是递增序列就lower_bound
+
+>如果是递减序列就手写二分
+
+## 区间dp
+1. 根据问题推出dp含义
+2. 根据规则写出dp的状态转移公式
+3. 处理边界问题
+> dp[i][j], dp[0][0], dp[i][0], dp[0][j], dp[i][i], dp[j][j]

src/8/26/P2758.cpp

@@ -0,0 +1,53 @@
+/*
+
+dp[i][j]=源串前i个字符匹配目标串前j个字符所需要的最小操作次数
+
+如果第i个字符==第j个字符
+dp[i][j]=dp[i-1][j-1]
+如果不相等
+dp[i][j]=min{
+    dp[i-1][j],
+    dp[i][j-1],
+    dp[i-1][j-1]
+}+1
+
+dp[i][j]=INT_MAX
+dp[0][0]=0
+dp[i][0]=i
+dp[0][j]=j
+*/
+#include <algorithm>
+#include <climits>
+#include <cstdint>
+#include <iostream>
+#include <istream>
+#include <string>
+#include <vector>
+using ll = int64_t;
+
+int main(){
+    std::iostream::sync_with_stdio(false);
+    std::cin.tie(nullptr);
+    std::string s1,s2;
+    std::cin>>s1>>s2;
+    ll n=s1.size(),m=s2.size();
+    s1=' '+s1;
+    s2=' '+s2;
+    std::vector<std::vector<ll>> dp(s1.size(),std::vector<ll>(s2.size(),INT_MAX));
+    for(ll i=0;i<dp.size();i++)dp[i][0]=i;
+    for(ll j=0;j<dp[0].size();j++)dp[0][j]=j;
+    for(ll i=1;i<=n;i++){
+        for(ll j=1;j<=m;j++){
+            if(s1[i]==s2[j]){
+                dp[i][j]=dp[i-1][j-1];
+            }else{
+                dp[i][j]=std::min({
+                    dp[i-1][j],
+                    dp[i][j-1],
+                    dp[i-1][j-1]
+                })+1;
+            }
+        }
+    }
+    std::cout<<dp[n][m]<<"\n";
+}

src/8/26/P2782.cpp

@@ -41,6 +41,7 @@ int main(){
         } else {
             *it = x;
         }
+        pa(f);
     }
     std::cout << f.size() << std::endl;
     return 0;

src/8/26/P5124.cpp

@@ -1,28 +1,27 @@
 #include <algorithm>
+#include <cstdint>
 #include <iostream>
 #include <istream>
-#include <numeric>
+#include <ostream>
 #include <vector>
-
 using ll = int64_t;
 
-#define pv(v)do{std::cout<<#v<<": "<<(v)<<"\n";}while(0)
-
 int main(){
     std::iostream::sync_with_stdio(false);
     std::cin.tie(nullptr);
-
     ll n,k;
     std::cin>>n>>k;
-    std::vector<ll> v(n);
-    for(ll&i:v){
-        std::cin>>i;
+    std::vector<ll> v(n+1),dp(n+1);
+    for(ll i=1;i<=n;i++){
+        std::cin>>v[i];
     }
-    std::sort(v.begin(),v.end());
-    ll teams=n/k,last=n%k;
-    ll teamssum = std::accumulate(v.rbegin(),v.rbegin()+teams,0);
-    // pv(teams);
-    // pv(teamssum);
-    // pv(last);
-    std::cout<<(last * (*(v.rbegin()+teams)) + teamssum*k)<<"\n";
+    for(ll i=1;i<=n;i++){
+        ll max=0;
+        for(ll len=1;len<=k&&len<=i;len++){
+            max=std::max(max,v[i-len+1]);
+            dp[i]=std::max(dp[i],max*len+dp[i-len]);
+        }
+    }
+    std::cout<<dp[n]<<std::endl;
+    _Exit(0);
 }

src/8/26/P5424.cpp

@@ -0,0 +1,68 @@
+/*
+
+dp[i][j]
+前i组蛇，用j个网的最小浪费空间
+
+dp[i][j]=min(dp[i][j],dp[m][j-1]+cnt[m+1][i])
+
+dp[i][j]=1e9
+dp[0][0]=0
+
+*/
+
+#include <algorithm>
+#include <cstdint>
+#include <cstdio>
+#include <iostream>
+#include <istream>
+#include <vector>
+
+using ll = int64_t;
+
+#define p2v(dp)do{\
+    for(ll i=0;i<dp.size();i++){\
+        for(ll j=0;j<dp[i].size();j++){\
+            printf("[%lld][%lld]=%lld\n",i,j,dp[i][j]);\
+        }\
+    }\
+}while(0)
+
+int main(){
+    std::iostream::sync_with_stdio(false);
+    std::cin.tie(nullptr);
+
+    ll n,m;
+    std::cin>>n>>m;
+    m++;
+    std::vector<ll> v(n+1);
+    std::vector<std::vector<ll>> cnt(n+1,std::vector<ll>(n+1));//cnt[i][j]=i-j用一个网浪费的空间
+    std::vector<std::vector<ll>> dp(n+1,std::vector<ll>(m+1, 1e9));
+    dp[0][0]=0;
+
+    for(ll i=1;i<=n;i++){
+        std::cin>>v[i];
+    }
+    for(ll i=1;i<=n;i++){
+        for(ll j=i;j<=n;j++){
+            ll nmax = *std::max_element(v.begin()+i,v.begin()+j+1);
+            ll ncnt=0;
+            for(ll m=i;m<=j;m++){
+                ncnt+=nmax-v[m];
+            }
+            cnt[i][j]=ncnt;
+        }
+    }
+    // printf("cnt\n");
+    // p2v(cnt);
+    for(ll i=1;i<=n;i++){
+        dp[i][1]=cnt[1][i];
+        for(ll j=2;j<=m&&j<=i;j++){
+            for(ll k=j-1;k<i;k++){
+                dp[i][j]=std::min(dp[i][j],dp[k][j-1]+cnt[k+1][i]);
+            }
+        }
+    }
+    // printf("dp\n");
+    // p2v(dp);
+    std::cout<<dp[n][m]<<"\n";
+}

src/8/26/P7414.cpp

@@ -0,0 +1,6 @@
+/*
+
+dp[i][j]=从第i个到第j个涂成指定颜色需要的次数
+
+
+*/