From 505d9e9c4d27d7b3e26ee1556d593d058e203bd6 Mon Sep 17 00:00:00 2001
From: Zengtudor <Zengtudor@outlook.com>
Date: Sat, 30 Aug 2025 16:42:52 +0800
Subject: [PATCH] =?UTF-8?q?feat:=20=E6=B7=BB=E5=8A=A0P1896.cpp=E8=A7=A3?=
 =?UTF-8?q?=E5=86=B3=E6=A3=8B=E7=9B=98=E6=94=BE=E7=BD=AE=E9=97=AE=E9=A2=98?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

实现动态规划算法解决棋盘上放置不互相攻击的国王问题。使用三维数组存储状态，通过位运算检查合法位置，并计算所有可能的放置方案总数。
---
 src/8/30/P1896.cpp | 78 ++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 78 insertions(+)
 create mode 100644 src/8/30/P1896.cpp

diff --git a/src/8/30/P1896.cpp b/src/8/30/P1896.cpp
new file mode 100644
index 0000000..98a4162
--- /dev/null
+++ b/src/8/30/P1896.cpp
@@ -0,0 +1,78 @@
+#include <cstdint>
+#include <cstdio>
+#include <cstdlib>
+#include <cstring>
+#include <iostream>
+#include <istream>
+#include <string>
+#include <vector>
+
+using ll = int64_t;
+
+ll n,K;
+std::vector<ll> line;
+std::vector<std::vector<std::vector<ll>>> dp;
+
+static inline constexpr ll get1n(ll x){
+    return __builtin_popcount(x);
+}
+
+int main(){
+    std::iostream::sync_with_stdio(false);
+    std::cin.tie(nullptr);
+
+    std::cin>>n>>K;
+    line.reserve(n*n+1);
+    line.push_back(0);
+    for(ll msk=0;msk<(1ll<<n);++msk){
+        for(ll i=1;i<=n;i++){
+            if((msk>>i)&1){
+                if((msk>>(i-1))&1 || msk>>(i+1)&1){
+                    goto nxt;
+                }
+            }
+        }
+        line.push_back(msk);
+        nxt:;
+    }
+    // for(ll i:line){
+    //     char buff[10];
+    //     memset(buff, 0, sizeof(buff));
+    //     itoa(i, buff, 2);
+    //     printf("%s\n",buff);
+    // }
+    dp.resize(n+1,std::vector<std::vector<ll>>(line.size(),std::vector<ll>(K+1)));
+
+    
+    for(ll j=1;j<line.size();++j){
+        ll num1 = get1n(line[j]);
+        dp[1][j][num1]=1;
+    }
+
+    for(ll i=2;i<=n;i++){
+        for(ll j=1;j<line.size();j++){
+            ll cur1n = get1n(line[j]);
+            for(ll k=1;k<line.size();k++){
+                ll k1n = get1n(line[k]);
+                if(k1n+cur1n>K)continue;
+                if(
+                    line[j] & line[k]
+                    || line[j] & (line[k]<<1)
+                    || line[j] & (line[k]>>1)
+                ){
+                    continue;
+                }else{
+                    for(ll x=cur1n;x<=K;K++){
+                        dp[i][j][x]+=dp[i-1][k][x-cur1n];
+                    }
+                }
+            }
+        }
+    }
+
+    ll ans=0;
+    for(ll j=1;j<line.size();j++){
+        ans+=dp[n][j][K];
+    }
+    std::cout<<ans<<"\n";
+}
\ No newline at end of file