From 52b2321a9664873ba638f9f4f4932385381ae73e Mon Sep 17 00:00:00 2001
From: Zengtudor <Zengtudor@outlook.com>
Date: Fri, 29 Aug 2025 14:17:10 +0800
Subject: [PATCH] =?UTF-8?q?feat:=20=E5=AE=9E=E7=8E=B0=E5=8A=A8=E6=80=81?=
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添加P4267.cpp文件，实现动态规划算法计算奶牛逃跑序列的最小修改次数。算法使用二维DP数组存储状态，并通过预处理计算区间修改成本。解决给定n头奶牛在不同逃跑次数下的最小不一致数量问题。
---
 src/8/28/P4267.cpp |  3 --
 src/8/29/P4267.cpp | 74 ++++++++++++++++++++++++++++++++++++++++++++++
 2 files changed, 74 insertions(+), 3 deletions(-)
 delete mode 100644 src/8/28/P4267.cpp
 create mode 100644 src/8/29/P4267.cpp

diff --git a/src/8/28/P4267.cpp b/src/8/28/P4267.cpp
deleted file mode 100644
index 294989d..0000000
--- a/src/8/28/P4267.cpp
+++ /dev/null
@@ -1,3 +0,0 @@
-int main(){
-    
-}
\ No newline at end of file
diff --git a/src/8/29/P4267.cpp b/src/8/29/P4267.cpp
new file mode 100644
index 0000000..cb548ee
--- /dev/null
+++ b/src/8/29/P4267.cpp
@@ -0,0 +1,74 @@
+/*
+
+dp[i][j]=前i个牛发生j次逃跑时序列不一致的数量最小数目
+
+dp[i][j]=min(dp[k][j-1]+cnt[k+1][i])
+
+dp[i][j]=inf
+
+dp[i][1]=cnt[1][i]
+
+*/
+
+#include <algorithm>
+#include <cstdint>
+#include <iostream>
+#include <istream>
+#include <vector>
+using ll = int64_t;
+
+const ll inf=1e9;
+
+#define p2v(arr)do{\
+    std::cout<<#arr<<":\n";\
+    for(auto&i:(arr)){\
+        for(auto&j:i){\
+            std::cout<<j<<", ";\
+        }\
+        std::cout<<"\n";\
+    }\
+}while(0)
+
+int main(){
+    std::iostream::sync_with_stdio(false);
+    std::cin.tie(nullptr);
+
+    ll n;
+    std::cin>>n;
+    std::vector<std::vector<ll>> dp(n+1,std::vector<ll>(n+1, inf)),
+            cnt(n+1,std::vector<ll>(n+1));
+    std::vector<ll> arr(n+1);
+    for(ll i=1;i<=n;i++){
+        std::cin>>arr[i];
+    }
+    /*
+    cost[i][j] 表示在区间 [i, j] 内，假设第 i 天发生逃跑，
+    并且在 (i, j] 区间内没有其他逃跑发生时，需要修改的记录数。
+    */
+    for(ll i=1;i<=n;i++){
+        ll mis=0;
+        for(ll j=i;j<=n;j++){
+            if(j-i!=arr[j]){
+                mis++;
+            }
+            cnt[i][j]=mis;
+        }
+    }
+    // p2v(cnt);
+    for(ll i=1;i<=n;i++){
+        dp[i][1]=cnt[1][i];
+    }
+    for(ll i=1;i<=n;i++){
+        for(ll j=2;j<=n;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]
+                );
+            }
+        }
+    }
+    for(ll j=1;j<=n;j++){
+        std::cout<<dp[n][j]<<"\n";
+    }
+}
\ No newline at end of file