feat: 添加P11454.cpp解题代码实现

实现了一个基于BFS的算法来解决P11454问题，包括输入处理、边界处理和结果计算。代码包含调试输出和性能优化措施。

src/10/25/P11454.cpp

@@ -0,0 +1,88 @@
+#include <cstdint>
+#include <cstdio>
+#include <deque>
+#include <iostream>
+#include <istream>
+#include <unordered_map>
+using ll = int64_t;
+
+const ll maxn=1007;
+ll n,q,r[maxn],c[maxn];
+char t[maxn];
+enum Dir{
+    U='U',D='D',R='R',L='L'
+};
+char map[maxn][maxn];
+bool vis[maxn][maxn];
+const int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
+char idxokdir[4]={U,L,D,R};
+int ans[int(2e5+7)];
+
+#define printf
+
+static inline ll bfs(ll x,ll y){
+    printf("bfs %lld,%lld\n",x,y);
+    ll res=0;
+    struct P{
+        ll x,y;
+    };
+    if(!vis[x][y])return 0;
+    std::deque<P> q;
+    q.emplace_back(x,y);
+    while (q.size()) {
+        auto[x,y]=q.front();
+        printf("get %lld %lld from q\n",x,y);
+        q.pop_front();
+        for(ll i=0;i<4;i++){
+            ll nx=x+dir[i][0],ny=y+dir[i][1];
+            if(nx<1||nx>n||ny<1||ny>n||vis[nx][ny]){
+                // printf("check %lld %lld not ok\n",nx,ny);
+                continue;
+            }
+            // printf("%lld %lld should be %c, but is %c\n",nx,ny,idxokdir[i],map[nx][ny]);
+            if(map[nx][ny]==idxokdir[i]||map[nx][ny]==0){
+                printf("ok %lld %lld\n",nx,ny);
+                q.emplace_back(nx,ny);
+                vis[nx][ny]=true;
+                res++;
+            }
+        }
+    }
+    printf("---out bfs\n");
+    return res;
+}
+
+int main(){
+    std::iostream::sync_with_stdio(false);
+    std::cin.tie(nullptr);
+
+    std::cin>>n>>q;
+    for(ll i=1;i<=q;i++){
+        std::cin>>r[i]>>c[i]>>t[i];
+        map[r[i]][c[i]]=t[i];
+    }
+    ll firstans=0;
+    for(ll i=1;i<=n;i++){
+        vis[0][i]=1;
+        vis[n+1][i]=1;
+        vis[i][0]=1;
+        vis[i][n+1]=1;
+        firstans+=bfs(0, i);
+        firstans+=bfs(n+1,i);
+        firstans+=bfs(i,0);
+        firstans+=bfs(i, n+1);
+    }
+    ans[q]=n*n-firstans;
+    printf("START for loop\n");
+    for(ll i=q;i>=2;i--){
+        printf("FOR i=%lld\n",i);
+        map[r[i]][c[i]]=0;
+        for(ll j=0;j<4;j++){
+            firstans+=bfs(r[i]+dir[j][0], c[i]+dir[j][1]);
+        }
+        ans[i-1]=n*n-firstans;
+    }
+    for(ll i=1;i<=q;i++){
+        std::cout<<ans[i]<<"\n";
+    }
+}