USACO 3.2 msquare 裸BFS

又是个裸BFS...

和西安网赛那道1006一样的，只不过加上了要记录方案。顺便复习map

记录方案直接在bfs队列的结点里加一个vector<int> opt，把从开头一直到当前结点的操作序列记下来

 /*
 PROB:msquare
 LANG:C++
 */

 #include <iostream>
 #include <vector>
 #include <algorithm>
 #include <map>
 #include <queue>
 #include <cstdio>
 using namespace std;

 struct node
 {
     vector<int> seq;
     int step;
     vector<int> opt;
     node(vector<int> x,int m,vector<int> o,int op):seq(x),step(m),opt(o)
     {
         opt.push_back(op);
     }
 };

 map<vector<int>,int> M;
 queue<node> Q;
 vector<int> a;      //tmp2
 vector<int> A;      //init
 vector<int> y;      //tmp1
 vector<int> B;      //goal
 vector<int> ans;
 vector<int> yy;
 int b[];

 bool same()
 {
     for (int i=;i<;i++)
         if (y[i]!=b[i])
             return false;
     return true;
 }

 int main()
 {
     freopen("msquare.in","r",stdin);
     freopen("msquare.out","w",stdout);

     B.clear();
     a.clear();
     A.clear();
     for (int i=;i<=;i++)
     {
         cin>>b[i];
         A.push_back(i+);
     }
     for (int i=;i>=;i--)
     {
         cin>>b[i];
         A.push_back(i+);
     }
     for (int i=;i<;i++)
         B.push_back(b[i]);

     //for (int i=0;i<8;i++)
     //    cout<<B[i]<<" "<<A[i]<<endl;

     //A[0]=1;     A[1]=2;     A[2]=3;     A[3]=4;
     //A[4]=8;     A[5]=7;     A[6]=6;     A[7]=5;
     Q.push(node(A,,a,));
     M.insert(pair<vector<int>,int>(A,));
     while (!Q.empty())
     {
         node tmp=Q.front();
         Q.pop();
         int st=tmp.step;
         //cout<<"step  "<<st<<endl;
         y=tmp.seq;
         yy=tmp.opt;
         if (same())
         {
             ans.clear();
             ans=tmp.opt;
             cout<<st<<endl;
             for (int i=;i<=st;i++)
             {
                 char ch=ans[i]+'A'-;
                 cout<<ch;
             }
             cout<<endl;
             break;
         }
         else
         {
             //opr1
             a=y;
             swap(a[],a[]);
             swap(a[],a[]);
             swap(a[],a[]);
             swap(a[],a[]);
             if (!M.count(a))
             {
                 M.insert(pair<vector<int>,int>(a,));
                 Q.push(node(a,st+,yy,));
             }
             //opr2
             a=y;
             int t1=a[],t2=a[];
             a[]=a[];  a[]=a[];
             a[]=a[];  a[]=a[];
             a[]=a[];  a[]=a[];
             a[]=t1;    a[]=t2;
             if (!M.count(a))
             {
                 M.insert(pair<vector<int>,int>(a,));
                 Q.push(node(a,st+,yy,));
             }
             //opr3
             a=y;
             int tm=a[];
             a[]=a[];  a[]=a[];  a[]=a[];  a[]=tm;
             if (!M.count(a))
             {
                 M.insert(pair<vector<int>,int>(a,));
                 Q.push(node(a,st+,yy,));
             }
         }
     }

     return ;
 }