URAL1034[Queens in peaceful positions]

先选三个来变，由于开始就很和谐，为了和谐，我们就将挑出来的三个棋子进行调整（按行或列来选），一共有两种调整方案（调整列或行）。

然后，在对角线判断一下就OK了。

CODE:

/*

PROGRAM: $PROGRAM

AUTHOR: Su Jiao

DATE: 2010-3-20

DESCRIPTION:

$DESCRIPTION

*/

#include <iostream>

using std::cin;

using std::cout;

#include <fstream>

using std::ifstream;

using std::ofstream;

#include <sstream>

using std::stringstream;

using std::endl;

#include <vector>

using std::vector;

#include <string>

using std::string;

#include <stack>

using std::stack;

#include <queue>

using std::queue;

#include <set>

using std::set;

#include <map>

using std::map;

using std::pair;

using std::make_pair;

#include <algorithm>

using std::sort;

#include <cassert>

//using std::assert;

 

class Application

{

      int N;

      vector<int> p;

      int total;

      public:

      Application()

      {

                   cin>>N;

                   p.resize(N);

                   for (int i=0;i<=N;i++)

                   {

                       int id,position;

                       cin>>id>>position;

                       p[id-1]=position-1;

                   }

      }

      bool can()

      {

           vector<bool> sum(N*2,false),delta(N*2,false);

           for (int i=0;i<N;i++)

           {

               if (sum[i+p[i]]) return false;

               else sum[i+p[i]]=true;

               if (delta[N+i-p[i]]) return false;

               else delta[N+i-p[i]]=true;

           }

           return true;

      }

      int run()

      {

          total=0;

          for (int c1=0;c1<N;c1++)

              for (int c2=c1+1;c2<N;c2++)

                  for (int c3=c2+1;c3<N;c3++)

              {

                  //backup

                  int b1=p[c1],b2=p[c2],b3=p[c3];

                  //CASE 1:

                  p[c1]=b3;

                  p[c2]=b1;

                  p[c3]=b2;

                  if (can()) total++;

                  //CASE 2:

                  p[c1]=b2;

                  p[c2]=b3;

                  p[c3]=b1;

                  if (can()) total++;

                  //recover

                  p[c1]=b1;

                  p[c2]=b2;

                  p[c3]=b3;

              }

          cout<<total<<endl;

          return 0;

      }

};

 

int main()

{

    Application app;

    return app.run();

}