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-8;
const double PI = atan(1.0)*4;
const int maxint = 2139062143;

struct nod{
    VI b;
    int d;
};

VI b;
int n;
nod an[1000000];
set<VI> st;
VI ans;


int BFS(int k)
{
    if (b == ans) return 0;
    st.erase(ALL(st));
    int l = 0, r = 0;
    VI ttt;
    nod tt; tt.d = 1;
    for (int i = 0; i+k <= n; ++ i){
        ttt = b;
        for (int j = i, t = i+k-1; 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 = 0; i+k <= n; ++ i){
            ttt = tmp.b;
            for (int j = i, t = i+k-1; 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 -1;
}

class SortingGame
{
    public:
        int fewestMoves(vector <int> B, int k){
            b = B; n = b.size();
            ans.clear();
            for (int i = 1; i <= n; ++ i)
                ans.PB(i);
            return BFS (k);
        }
        
// BEGIN CUT HERE
    public:
    void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); if ((Case == -1) || (Case == 4)) 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 << "	Expected: "" << Expected << '"' << endl; cerr << "	Received: "" << Received << '"' << endl; } }
    void test_case_0() { int Arr0[] = {1,2,3}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 3; int Arg2 = 0; verify_case(0, Arg2, fewestMoves(Arg0, Arg1)); }
    void test_case_1() { int Arr0[] = {3,2,1}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 3; int Arg2 = 1; verify_case(1, Arg2, fewestMoves(Arg0, Arg1)); }
    void test_case_2() { int Arr0[] = {5,4,3,2,1}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 2; int Arg2 = 10; verify_case(2, Arg2, fewestMoves(Arg0, Arg1)); }
    void test_case_3() { int Arr0[] = {3,2,4,1,5}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 4; int Arg2 = -1; verify_case(3, Arg2, fewestMoves(Arg0, Arg1)); }
    void test_case_4() { int Arr0[] = {7,2,1,6,8,4,3,5}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 4; int Arg2 = 7; verify_case(4, Arg2, fewestMoves(Arg0, Arg1)); }

// END CUT HERE

};

// BEGIN CUT HERE
int main()
{
//    freopen( "a.out" , "w" , stdout );    
    SortingGame ___test;
    ___test.run_test(-1);
       return 0;
}
// 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] = 0;
 
    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 = 1; i < N; i++)
        if (u[i-1] > u[i]) break;
      if (i == N) return d;
 
      for (i = 0; i + k <= N; i++) {
        VI w = u;
        reverse(w.begin() + i, w.begin() + i + k);
        if (dist.count(w) == 0) {
          dist[w] = d + 1;
          Q.push(w);
        }
      }
    }
 
    return -1;    
  }
};