SRM 397(1-250pt)

题意：对于一个长度n的数列（由1-n组成，n <= 8），每次操作可以reverse k个连续的数。问最少多少次操作可以将该数列转化成递增的数列。

解法：就是一个BFS。只是由于最开始学习BFS时写法很复杂，所以直到现在也觉得BFS不好写。然后就在想还有没有别的方法，然后时间就浪费了...最后还是写了个比较冗杂的BFS，150+score。看了官方题解的BFS，真是学习了...一方面是BFS的写法搓，另一方面是不会用C++内置的reverse函数。

tag:BFS

我的代码：

 // BEGIN CUT HERE
 /*
  * Author:  plum rain
  * score :
  */
 /*

  */
 // END CUT HERE
 #line 11 "SortingGame.cpp"
 #include <sstream>
 #include <stdexcept>
 #include <functional>
 #include <iomanip>
 #include <numeric>
 #include <fstream>
 #include <cctype>
 #include <iostream>
 #include <cstdio>
 #include <vector>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <cstdlib>
 #include <set>
 #include <queue>
 #include <bitset>
 #include <list>
 #include <string>
 #include <utility>
 #include <map>
 #include <ctime>
 #include <stack>

 using namespace std;

 #define CLR(x) memset(x, 0, sizeof(x))
 #define CLR1(x) memset(x, -1, sizeof(x))
 #define PB push_back
 #define SZ(v) ((int)(v).size())
 #define ALL(t) t.begin(),t.end()
 #define zero(x) (((x)>0?(x):-(x))<eps)
 #define out(x) cout<<#x<<":"<<(x)<<endl
 #define tst(a) cout<<#a<<endl
 #define CINBEQUICKER std::ios::sync_with_stdio(false)

 typedef vector<int> VI;
 typedef vector<string> VS;
 typedef vector<double> VD;
 typedef pair<int, int> pii;
 typedef long long int64;

 const double eps = 1e-;
 const double PI = atan(1.0)*;
 const int maxint = ;

 struct nod{
     VI b;
     int d;
 };

 VI b;
 int n;
 nod an[];
 set<VI> st;
 VI ans;

 int BFS(int k)
 {
     if (b == ans) return ;
     st.erase(ALL(st));
     int l = , r = ;
     VI ttt;
     nod tt; tt.d = ;
     for (int i = ; i+k <= n; ++ i){
         ttt = b;
         for (int j = i, t = i+k-; j < t; ++ j, -- t)
             swap(ttt[j], ttt[t]);
         tt.b = ttt;
         if (!st.count(ttt)){
             an[r++] = tt;
             st.insert(ttt);
         }
     }

     while (l < r){
         nod tmp = an[l++];
         if (tmp.b == ans) return tmp.d;
         ++ tmp.d;
         for (int i = ; i+k <= n; ++ i){
             ttt = tmp.b;
             for (int j = i, t = i+k-; j < t; ++ j, -- t)
                 swap(ttt[j], ttt[t]);
             if (!st.count(ttt)){
                 an[r].b = ttt;
                 an[r++].d = tmp.d;
                 st.insert(ttt);
             }
         }
     }
     return -;
 }

 class SortingGame
 {
     public:
         int fewestMoves(vector <int> B, int k){
             b = B; n = b.size();
             ans.clear();
             for (int i = ; i <= n; ++ i)
                 ans.PB(i);
             return BFS (k);
         }

 // BEGIN CUT HERE
     public:
     void run_test(int Case) { if ((Case == -) || (Case == )) test_case_0(); if ((Case == -) || (Case == )) test_case_1(); if ((Case == -) || (Case == )) test_case_2(); if ((Case == -) || (Case == )) test_case_3(); if ((Case == -) || (Case == )) test_case_4(); }
     private:
     template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
     void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
     void test_case_0() { int Arr0[] = {,,}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; verify_case(, Arg2, fewestMoves(Arg0, Arg1)); }
     void test_case_1() { int Arr0[] = {,,}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; verify_case(, Arg2, fewestMoves(Arg0, Arg1)); }
     void test_case_2() { int Arr0[] = {,,,,}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; verify_case(, Arg2, fewestMoves(Arg0, Arg1)); }
     void test_case_3() { int Arr0[] = {,,,,}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = -; verify_case(, Arg2, fewestMoves(Arg0, Arg1)); }
     void test_case_4() { int Arr0[] = {,,,,,,,}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; verify_case(, Arg2, fewestMoves(Arg0, Arg1)); }

 // END CUT HERE

 };

 // BEGIN CUT HERE
 int main()
 {
 //    freopen( "a.out" , "w" , stdout );
     SortingGame ___test;
     ___test.run_test(-);
        return ;
 }
 // END CUT HERE

官方题解推荐代码：

 #include <algorithm>
 #include <iostream>
 #include <sstream>
 #include <vector>
 #include <cmath>
 #include <map>
 #include <queue>
 using namespace std;
 typedef vector<int> VI;

 struct SortingGame {
   int fewestMoves(vector<int> board, int k) {
     map<VI, int> dist;
     dist[board] = ;

     queue<VI> Q;
     Q.push(board);

     int i, N = board.size();

     while (!Q.empty()) {
       VI u = Q.front(); Q.pop();
       int d = dist[u];

       for (i = ; i < N; i++)
         if (u[i-] > u[i]) break;
       if (i == N) return d;

       for (i = ; i + k <= N; i++) {
         VI w = u;
         reverse(w.begin() + i, w.begin() + i + k);
         if (dist.count(w) == ) {
           dist[w] = d + ;
           Q.push(w);
         }
       }
     }

     return -;
   }
 };