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Pku 1065 Wooden Sticks

Darren — Thu, 30 Oct 2008 12:55:00 GMT

#include < stdio.h >
#include < stdlib.h >
#include < string .h >

#define N 5001

struct Node
{
     int length;
     int weight;
};

int n;
Node d[N];
bool visite[N];

int cmp( const void * a, const void * b )
{
    Node * ta = (Node * )a;
    Node * tb = (Node * )b;

     if ( ta -> length == tb -> length ) return ta -> weight - tb -> weight;

     return ta -> length - tb -> length;
}

int main()
{
     int test;
    scanf( " %d " , & test);

     while ( test -- )
    {
        scanf( " %d " , & n );
         for ( int i = 0 ; i < n; ++ i )
        scanf( " %d%d " ,      & d[i].length, & d[i].weight );

        qsort( d, n, sizeof (d[ 0 ]), cmp );
        memset( visite, false , sizeof (visite) );

         int b = d[ 0 ].weight;
         int num = 0 ;
         int k = 0 ;

         while ( true )
        {
             bool ok = false ;

             if ( k == - 1 ) break ;
            b = d[k].weight;   num ++ ;

             int be = k;  k = - 1 ;

             for ( int i = be; i < n; ++ i )
            {
                 if ( d[i].weight >= b && ! visite[i] )
                {
                    visite[i] = true ;
                    b = d[i].weight;
                }
                 else if ( ! ok && ! visite[i] )   {   k = i;  ok = true ;  }
            }
        }

        printf( " %d\n " , num );
    }

     return 0 ;
}

Darren 2008-10-30 20:55 发表评论

Pku 2488 A Knight's Journey

Darren — Tue, 14 Oct 2008 04:00:00 GMT

#include < stdio.h >
#include < stdlib.h >

#define pause system("pause")

int r, c;
bool ok = false ;

bool visite[ 100 ][ 100 ];
int   px[ 50 ], py[ 50 ];

int yy[ 8 ] = { - 1 , 1 , - 2 , 2 , - 2 , 2 , - 1 , 1 };
int xx[ 8 ] = { - 2 , - 2 , - 1 , - 1 ,   1 , 1 ,   2 , 2 };

bool isok( int x, int y )
{
     return x >= 1 && x <= r && y >= 1 && y <= c;
}

void dfs( int x, int y, int length )
{
    px[length] = x, py[length] = y;

     if ( length == r * c )
    {
        ok = true ;
         return ;
    }

     for ( int i = 0 ; i < 8 ; ++ i )
    {
         int tx = xx[i] + x, ty = yy[i] + y;

         if ( isok(tx,ty) && ! visite[tx][ty] && ! ok )
        {
            visite[tx][ty] = true ;

            dfs( tx, ty, length + 1 );

            visite[tx][ty] = false ;
        }
    }
}

int main()
{
     int test;
    scanf( " %d " , & test);

     for ( int t = 1 ; t <= test; ++ t )
    {
        scanf( " %d%d " , & c, & r);

        visite[ 1 ][ 1 ] = true ; ok = false ;
        dfs( 1 , 1 , 1 );

        printf( " Scenario #%d:\n " , t );

         if ( ok )
        {
             for ( int i = 1 ; i <= r * c; ++ i )
            {
                printf( " %c " , px[i] + ' A ' - 1 );
                printf( " %d " , py[i] );
            }
            printf( " \n " );
        }
         else puts( " impossible " );

         if ( t < test ) puts( "" );
    }

     return 0 ;
}

Darren 2008-10-14 12:00 发表评论

PKU 1018 Communication System

Darren — Wed, 01 Oct 2008 14:06:00 GMT

#include < iostream >
#include < set >
#include < limits >

using namespace std;

struct Node
{
     int broad, price;
    Node() {}
    Node( int & a, int & b)
        :broad(a),price(b)
     {}
} ;

Node data[ 100 ][ 100 ];
int   length[ 100 ];

int main()
{
     int test;
    scanf( " %d " , & test);

     while ( test -- )
     {
         int n;
         set < int > br;

        scanf( " %d " , & n);

         for ( int i = 0 ; i < n; ++ i )
         {
             int d,x,y;
            scanf( " %d " , & d);

            length[i] = d;
             for ( int j = 0 ; j < d; ++ j )
             {
                scanf( " %d%d " , & x, & y);
                data[i][j] = Node(x,y);

                br.insert(x);     //    Save all broad , use set to avoid repeating
            }
        }

         set < int > ::iterator pos;
         double max = numeric_limits < double > ::min();
         for ( pos = br.begin (); pos != br.end (); ++ pos )    //    for every broad
         {
             int    t = * pos;
             int    total = 0 ;
             bool   isok = false ;

             for ( int i = 0 ; i < n; ++ i )    //   n devices
             {
                 int M = INT_MAX;

                 for ( int j = 0 ; j < length[i]; ++ j )
                     if ( data[i][j].broad >= t && data[i][j].price < M )
                        M = data[i][j].price;       //    find  the min price

                 if ( M == INT_MAX )
                 {
                    isok = true ;
                     break ;
                }
                total += M;
            }

             if ( isok ) continue ;

             if ( ( double )t / ( double )total > max ) max = ( double )t / ( double )total;    //   update
        }

        printf( " %.3lf\n " , max );
    }

     return 0 ;
}

Darren 2008-10-01 22:06 发表评论

pku 1017 Packets

Darren — Wed, 01 Oct 2008 11:42:00 GMT

摘要: #include < iostream > using namespace std; int d[ 6 ]; int main() ... 阅读全文

Darren 2008-10-01 19:42 发表评论

PKU 1014 Dividing

Darren — Wed, 01 Oct 2008 07:38:00 GMT

#include < iostream >

using namespace std;

int   d[ 6 ];
bool isok = false ;
int   total = 0 ;

void Try( int i, int total )
{
     if ( total == 0 )
     {
        isok = true ;
         return ;
    }

     if ( i < 0 ) return ;

     for ( int j = d[i]; j >= 0 ; j -- )
     {
         if ( isok ) break ;
         if ( total - j * (i + 1 ) >= 0 && ! isok )
            Try( i - 1 , total - j * (i + 1 ) );
    }
}

int main()
{
     int num = 1 ;
     while ( scanf( " %d%d%d%d%d%d " , d + 0 , d + 1 , d + 2 , d + 3 , d + 4 , d + 5 ),
           d[ 0 ] + d[ 1 ] + d[ 2 ] + d[ 3 ] + d[ 4 ] + d[ 5 ] != 0 )
     {
        total = 0 ;
         for ( int i = 0 ; i < 6 ; ++ i )
         {
            d[i] %= 60 ;
            total += (i + 1 ) * d[i];
        }

         if ( total & 1 )
         {
            printf( " Collection #%d:\n " ,num ++ );
            printf( " Can't be divided.\n " );
            printf( " \n " );
             continue ;
        }

        isok = false ;
        Try( 5 , total / 2 );

        printf( " Collection #%d:\n " ,num ++ );

         if ( ! isok )
            printf( " Can't be divided.\n " );
         else
            printf( " Can be divided.\n " );
        printf( " \n " );
    }

     return 0 ;
}

Darren 2008-10-01 15:38 发表评论

Open the Lock

Darren — Tue, 19 Aug 2008 08:38:00 GMT

摘要: Problem description Now an emergent task for you is to open a password lock. The password is consisted of four digits. Each digit is numbered from 1 to 9. Each time, you can add or minus 1 to any ... 阅读全文

Darren 2008-08-19 16:38 发表评论

Sudoku

Darren — Tue, 19 Aug 2008 00:14:00 GMT

摘要: 出处： http://acm.hnu.cn:8080/online/?action=problem&type=show&id=11203&courseid=0 Sudoku ... 阅读全文

Darren 2008-08-19 08:14 发表评论

Change binary string

Darren — Mon, 18 Aug 2008 12:14:00 GMT

摘要: Problem description There are two binary strings, their length is 8, you should change the first to the second through several approach. You ne... 阅读全文

Darren 2008-08-18 20:14 发表评论