C++博客-Darren-随笔分类-动态规划

pku 1691 Painting A Board 状态压缩DP

Darren — Thu, 08 Oct 2009 06:54:00 GMT

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

int const inf = 1 << 29 ;
int n;
struct Point {
     int Lx, Ly, Rx, Ry;
} ;

Point pnt[ 16 ];
int   need[ 16 ], dp[ 1 << 15 ][ 15 ], color[ 16 ];

void init() {
     for ( int i = 0 ; i <= n; ++ i ) need[i] = 0 ;

     for ( int i = 0 ; i < n; ++ i ) {
         for ( int j = 0 ; j < n; ++ j ) {
             if ( i == j ) continue ;

             if ( pnt[j].Ry <= pnt[i].Ly && pnt[j].Lx < pnt[i].Rx ) {
                need[i] |= ( 1 << j);
            }
        }
    }
}

void solve() {
     for ( int i = 0 ; i < ( 1 << n); ++ i )
         for ( int j = 0 ; j < n; ++ j ) dp[i][j] = inf;

    dp[ 0 ][ 0 ] = 0 ;
     for ( int i = 0 ; i < n; ++ i )
         if ( need[i] == 0 ) dp[ 1 << i][i] = 1 ;

     for ( int s = 0 ; s < ( 1 << n); ++ s ) {
         for ( int i = 0 ; i < n; ++ i ) {
             if ( ( s & ( 1 << i) ) == 0 ) continue ;

             for ( int j = 0 ; j < n; ++ j ) {
                 if ( s & ( 1 << j) ) continue ;
                 if ( (s & need[j]) != need[j] ) continue ;

                 if ( dp[s][i] + ( color[i] != color[j] ) < dp[s | ( 1 << j)][j] ) {
                    dp[s | ( 1 << j)][j] = dp[s][i] + ( color[i] != color[j] );
                }
            }
        }
    }

     int ans = inf;
     for ( int i = 0 ; i < n; ++ i )
         if ( dp[( 1 << n) - 1 ][i] < ans ) ans = dp[( 1 << n) - 1 ][i];

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

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

     while ( test -- ) {
        scanf( " %d " , & n );

         for ( int i = 0 ; i < n; ++ i ) {
            scanf( " %d%d%d%d " , & pnt[i].Ly, & pnt[i].Lx, & pnt[i].Ry,   & pnt[i].Rx );
            scanf( " %d " , color + i );
        }

        init();
        solve();
    }

     return 0 ;
}

Darren 2009-10-08 14:54 发表评论

Pku 1276 Cash Machine

Darren — Sun, 19 Jul 2009 05:44:00 GMT

#include < iostream >

using namespace std;

#define N 100001

int nk[ 11 ], dk[ 11 ], n, m, ans[N], num[N];

int main(){
     while ( scanf( " %d%d " , & m, & n ) != EOF ){
         for ( int i = 0 ; i < n; ++ i ) scanf( " %d%d " , nk + i, dk + i );
         for ( int i = 0 ; i <= m; ++ i ) ans[i] = 0 ; ans[ 0 ] = 1 ;

         for ( int i = 0 ; i < n; ++ i ){
             for ( int j = 0 ; j <= m; ++ j ) num[j] = 0 ;

             for ( int j = dk[i]; j <= m; ++ j )
             if ( ! ans[j] && ans[j - dk[i]] && num[j - dk[i]] < nk[i] ){
                num[j] = num[j - dk[i]] + 1 ;
                ans[j] = 1 ; }
        }

         int res = m;
         while ( ans[res] == 0 && res >= 1 ) res -- ;
        printf( " %d\n " , res );
    }

     return 0 ;
}

Darren 2009-07-19 13:44 发表评论

Pku 1651 Multiplication Puzzle

Darren — Mon, 03 Nov 2008 09:56:00 GMT

#include < stdio.h >

int      n;
int      r[ 110 ][ 110 ];
int      p[ 110 ];

int main()
{
     while ( scanf( " %d " , & n) != EOF )
     {
         for ( int i = 0 ; i < n; ++ i )
         {
            scanf( " %d " , & p[i] );
            r[i][i] = 0 ; r[i + 1 ][i] = 0 ; r[i][i + 1 ] = 0 ;
        }

         for ( int d = 2 ; d < n; ++ d )
             for ( int i = 0 ; i < n - d; ++ i )
             {
                 int j = i + d;

                r[i][j] = r[i + 1 ][j] + p[i] * p[i + 1 ] * p[j];

                 for ( int k = i + 1 ; k < j; ++ k )
                 {
                     int t = r[i][k] + r[k][j] + p[i] * p[k] * p[j];

                     if ( t < r[i][j] ) r[i][j] = t;
                }
            }

        printf( " %d\n " , r[ 0 ][n - 1 ] );
    }

     return 0 ;
}

Darren 2008-11-03 17:56 发表评论

矩阵连乘

Darren — Mon, 03 Nov 2008 09:05:00 GMT

#include < stdio.h >

struct   Matrix
{
     int r,c;
} ;

int      n;
int      r[ 110 ][ 110 ];
Matrix  p[ 110 ];    //    The Matrix with row and column

/*

6
30 35 35 15 15 5 5 10 10 20 20 25

*/

int main()
{
     while ( scanf( " %d " , & n) != EOF )
     {
         for ( int i = 0 ; i < n; ++ i )
         {
            scanf( " %d%d " , & p[i].r, & p[i].c );
            r[i][i] = 0 ;
        }

         for ( int d = 1 ; d < n; ++ d )
             for ( int i = 0 ; i < n - d; ++ i )
             {
                 int j = i + d;

                r[i][j] = r[i + 1 ][j] + p[i].r * p[i].c * p[j].c;

                 for ( int k = i + 1 ; k < j; ++ k )
                 {
                     int t = r[i][k] + r[k + 1 ][j] + p[i].r * p[k].c * p[j].c;

                     if ( t < r[i][j] ) r[i][j] = t;
                }
            }

        printf( " %d\n " , r[ 0 ][n - 1 ] );
    }

     return 0 ;
}

Darren 2008-11-03 17:05 发表评论

Pku 1023 The Fun Number System

Darren — Thu, 30 Oct 2008 03:57:00 GMT

#include < stdio.h >

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

     while ( test -- )
     {
        __int64   len, data;
         char       str[ 100 ];
         int        result[ 200 ], t = 0 ;

        scanf( " %I64d%s%I64d " , & len,str, & data );

         for ( int i = len - 1 ; i >= 0 ; i -- )
         {
             int a = ( data % 2 + 2 ) % 2 ;
            result[t ++ ] = a;

             if ( str[i] == ' n ' )  data = ( data + a ) / 2 ;
             else                 data = ( data - a ) / 2 ;
        }

         if ( data == 0 )
         {
             for ( int i = t - 1 ; i >= 0 ; i -- )
            printf( " %d " , result[i] );

            printf( " \n " );
        }
         else   printf( " Impossible\n " );
    }

     return 0 ;
}

Darren 2008-10-30 11:57 发表评论

Pku 1032 Parliament

Darren — Wed, 29 Oct 2008 02:26:00 GMT

#include < stdio.h >

int result[ 500 ];

int main()
{
     int n;

     while ( scanf( " %d " , & n) != EOF )
     {
         int sum = 0 ;
         int len = 0 ;
         int value = 2 ;

         while ( true ) {
            sum += value;
            result[len ++ ] = value ++ ;
             if ( sum + value > n ) break ;
        }
        value -- ;
         int left = n - sum;
         if ( value == left )
         {
             for ( int i = 0 ; i < len; ++ i ) result[i] ++ ;
            result[len - 1 ] ++ ;
        }
         else for ( int i = len - 1 ; i >= 0 , left > 0 ; i -- , left -- ) result[i] ++ ;

         for ( int i = 0 ; i < len; ++ i )
         {
             if ( i == 0 ) printf( " %d " , result[i] );
             else         printf( " %d " , result[i] );
        }
        printf( " \n " );
    }

     return 0 ;
}

做法就是求出以2起始的最大连续自然数序列之和sum，使得sum的值不超过输入数n，
然后分情况讨论：

设此最大序列为2、3、……、w，则：

1。若剩余值（n-sum）等于w，则最后输出序列为：3、4、……、w、w+2，即将原最大序列每项加1，再将最后剩余的一个1加到最后一项上。

2。若剩余值（n-sum）小于w，则从序列的最大项i开始，从大到小依次将每项加1，直到剩余值用完。

Darren 2008-10-29 10:26 发表评论

Pku 1046 Color Me Less

Darren — Tue, 28 Oct 2008 14:08:00 GMT

#include < stdio.h >
#include < math.h >

struct Point
{
     double x, y, z;
} ;

Point  d[ 16 ];

double dis( Point const & a, Point const & b )
{
     return sqrt( ( a.x - b.x ) * (a.x - b.x ) + ( a.y - b.y ) * (a.y - b.y ) + (a.z - b.z) * (a.z - b.z) );
}


int main()
{
     for ( int i = 0 ; i < 16 ; ++ i )  scanf( " %lf%lf%lf " , & d[i].x, & d[i].y, & d[i].z );
    Point temp;

     while ( scanf( " %lf%lf%lf " , & temp.x, & temp.y, & temp.z), temp.x != - 1 && temp.y != - 1 && temp.z != - 1 )
     {
         double min = 250 * 250 + 250 * 250 + 250 * 250 + 50 ;
         int     k = - 1 ;

         for ( int i = 0 ; i < 16 ; ++ i )
         {
             if ( dis( d[i], temp ) < min )
             {
                min = dis( d[i], temp );
                k = i;
            }
        }

        printf( " (%.0lf,%.0lf,%.0lf) maps to (%.0lf,%.0lf,%.0lf)\n " ,  temp.x, temp.y, temp.z, d[k].x, d[k].y, d[k].z );
    }

     return 0 ;
}

Darren 2008-10-28 22:08 发表评论

Pku 1080 Human Gene Functions

Darren — Tue, 28 Oct 2008 10:34:00 GMT

#include < stdio.h >

int m, n;
char a[ 110 ], b[ 110 ];
int   result[ 110 ][ 110 ];
int   value[ 5 ][ 5 ] = {    { 5 , - 1 , - 2 , - 1 , - 3 } ,
                       { - 1 , 5 , - 3 , - 2 , - 4 } ,
                       { - 2 , - 3 , 5 , - 2 , - 2 } ,
                       { - 1 , - 2 , - 2 , 5 , - 1 } ,
                       { - 3 , - 4 , - 2 , - 1 , - 1000 } } ;

int getid( char ch )
{
     switch ( ch )
     {
         case ' A ' :   return 0 ;
         case ' C ' :   return 1 ;
         case ' G ' :   return 2 ;
         case ' T ' :   return 3 ;
         case ' - ' :   return 4 ;
    }

     return - 1 ;
}

int solve()
{
    result[ 0 ][ 0 ] = 0 ;
     for ( int i = 0 ; i < m; ++ i ) result[i + 1 ][ 0 ] = result[i][ 0 ] + value[ getid( a[i] ) ][ 4 ];
     for ( int i = 0 ; i < n; ++ i ) result[ 0 ][i + 1 ] = result[ 0 ][i] + value[ 4 ][ getid( b[i] ) ];

     for ( int i = 0 ; i < m; ++ i )
         for ( int j = 0 ; j < n; ++ j )
         {
            result[i + 1 ][j + 1 ] = result[i][j] + value[ getid( a[i] ) ][ getid( b[j] ) ];

             if ( result[i + 1 ][j] + value[ 4 ][ getid( b[j] ) ] > result[i + 1 ][j + 1 ] )
            result[i + 1 ][j + 1 ] = result[i + 1 ][j] + value[ 4 ][ getid( b[j] ) ];

             if ( result[i][j + 1 ] + value[ getid( a[i] ) ][ 4 ] > result[i + 1 ][j + 1 ] )
            result[i + 1 ][j + 1 ] = result[i][j + 1 ] + value[ getid( a[i] ) ][ 4 ];
        }

     return result[m][n];
}

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

     while ( test -- )
     {
        scanf( " %d " , & m); scanf( " %s " , a );
        scanf( " %d " , & n); scanf( " %s " , b );

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

     return 0 ;
}

Darren 2008-10-28 18:34 发表评论

Pku 3096 Surprising Strings

Darren — Fri, 10 Oct 2008 04:18:00 GMT

#include < map >
#include < cstdio >

using namespace std;

struct Node
{
     char   a, b;
    Node() {}
    Node( char x, char y):a(x),b(y) {}
} ;

bool operator < ( Node x, Node y )
{
     if ( x.a == y.a )
     return x.b < y.b;
     return x.a < y.a;
}

int main()
{
     char str[ 80 ];

     while ( gets(str), str[ 0 ] != ' * ' )
     {
         bool b = true ;

         for ( int t = 1 ; t < strlen(str); ++ t )
         {
            map < Node, int > d;
             bool isok = true ;

             for ( int i = 0 ; i < strlen(str); ++ i )
             {
                 if ( (i + t) < strlen(str) && d[ Node( str[i], str[t + i] ) ] == 0 )
                     d[ Node( str[i], str[t + i] ) ] = 1 ;
                 else if ( (i + t) < strlen(str) && d[ Node( str[i],str[t + i]) ] == 1 )
                 {
                     isok = false ;
                      break ;
                }
            }

             if ( ! isok )
             {
                b = false ;
                 break ;
            }
        }

         if ( b )   printf( " %s is surprising.\n " ,str);
         else       printf( " %s is NOT surprising.\n " ,str);
    }

     return 0 ;
}

Darren 2008-10-10 12:18 发表评论

Pku 1745 Divisibility

Darren — Thu, 09 Oct 2008 04:37:00 GMT

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

int   n, k;
int   d[ 10001 ];
bool result[ 10001 ][ 101 ];

int main()
{
     while ( scanf( " %d%d " , & n, & k) != EOF )
     {
         for ( int i = 1 ; i <= n; ++ i ) scanf( " %d " , & d[i] );

        memset( result, false , sizeof (result) );

         int t = d[ 1 ] > 0 ? ( d[ 1 ] % k ): ( (k - ( - d[ 1 ] % k) ) % k );
        result[ 1 ][t] = true ;

         for ( int i = 2 ; i <= n; ++ i )
         {
             int t1 = ( d[i] > 0 ) ? (d[i] % k):( - d[i] % k);
             int t2 = (k - t1) % k;

             for ( int j = 0 ; j < 101 ; ++ j )
                 if ( result[i - 1 ][j] )
                 {
                    result[i][ (j + t1) % k ] = true ;
                    result[i][ (j + t2) % k ] = true ;
                }
         }

          if ( result[n][ 0 ] ) printf( " Divisible\n " );
          else                 printf( " Not divisible\n " );
    }

     return 0 ;
}

Darren 2008-10-09 12:37 发表评论

Pku 1837 Balance

Darren — Thu, 09 Oct 2008 03:14:00 GMT

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

int   mid = 5000 ;
int   result[ 21 ][ 10000 ];
int   C,G;
int   pos[ 21 ];
int   value[ 21 ];

int main()
{
     while ( scanf( " %d%d " , & C, & G) != EOF )
     {
         for ( int i = 1 ; i <= C; ++ i )  scanf( " %d " , & pos[i] );
         for ( int i = 1 ; i <= G; ++ i )  scanf( " %d " , & value[i] );

        memset( result, 0 , sizeof (result ) );

         for ( int i = 1 ; i <= C; ++ i )
        result[ 1 ][ mid + pos[i] * value[ 1 ] ] ++ ;

         for ( int i = 2 ; i <= G; ++ i )
             for ( int j = 1 ; j <= C; ++ j )
                 for ( int k = 0 ; k < 10000 ; ++ k )
                    if ( result[i - 1 ][k] > 0 )
                   result[i][ k + value[i] * pos[j] ] += result[ i - 1 ][k];

        printf( " %d\n " , result[G][mid] );
    }

     return 0 ;
}

Darren 2008-10-09 11:14 发表评论

Pku_1088 滑雪

Darren — Fri, 03 Oct 2008 04:29:00 GMT

#include < iostream >
#include < limits >

int data[ 100 ][ 100 ];
int resu[ 100 ][ 100 ];
int x[ 4 ] = { - 1 , 0 , 1 , 0 } ;
int y[ 4 ] = {   0 , - 1 , 0 , 1 } ;
int r,c;

bool isok( int i, int j )
{
     return ( i >= 0 && i < r && j >= 0 && j < c );
}

int DP( int i, int j )
{
     int max = 0 ;

     if ( resu[i][j] != 0 ) return resu[i][j];

     for ( int t = 0 ; t < 4 ; ++ t )
     {

         if ( isok( i + x[t], j + y[t] ) && data[ i + x[t] ][ j + y[t] ] < data[i][j] )
         {
             int m = DP( i + x[t], j + y[t] );

             if ( m > max ) max = m;
        }
    }

     return max + 1 ;
}

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

     for ( int i = 0 ; i < r; ++ i )
         for ( int j = 0 ; j < c; ++ j )
            scanf( " %d " , & data[i][j] );

    memset( resu, 0 , sizeof (resu) );

     int max = INT_MIN;
     for ( int i = 0 ; i < r; ++ i )
         for ( int j = 0 ; j < c; ++ j )
         {
            resu[i][j] = DP(i,j);

             if ( max < resu[i][j] ) max = resu[i][j];
        }

    printf( " %d\n " , max );

     return 0 ;
}

Darren 2008-10-03 12:29 发表评论

拦截导弹

Darren — Tue, 19 Aug 2008 08:32:00 GMT

Problem description

某国为了防御敌国的导弹袭击，发展出一种导弹拦截系统。但是这种导弹拦截系统有一个缺陷：虽然它的第一发炮弹能够到达任意的高度，但是以后每一发炮弹都不能高于前一发的高度。某天，雷达捕捉到敌国的导弹来袭。由于该系统还在试用阶段，所以只有一套系统，因此有可能不能拦截所有的导弹。

Input
输入数据为导弹依次飞来的高度，所有高度值均为不大于30000的正整数。

Output
输出只有一行是这套系统最多能拦截的导弹数和要拦截所有导弹最少要配备这种导弹拦截系统的套数。两个数据之间用一个空格隔开.

Sample Input
389 207 155 300 299 170 158 65

Sample Output
6 2

第一问显然是求最长不下降序列长度
第二问求最长上升序列长度

1 #include < iostream >
2 #include < vector >
3
4 using namespace std;
5
6 int main()
7 {
8     vector < int >   a;
9      int     data;
10
11      while ( cin >> data )
12         a.push_back( data );
13
14      int    len = static_cast < int > ( a.size () );
15      int *   value = new int [len];
16      int    max, t = 0 ;
17
18      for ( int i = 0 ; i < len; ++ i )
19         value[i] = 1 ;
20
21      for ( int i = 0 ; i < len; ++ i )
22      {
23         max = 0 ;
24         t = 0 ;
25
26          for ( int j = 0 ; j < i; ++ j )
27              if ( a[i] < a[j] && value[j] > max )
28              {
29                 t = 1 ;
30                 max = value[j];
31             }
32
33          if ( t != 0 )
34             value[i] = max + 1 ;
35     }
36
37     max = 1 ;
38      for ( int i = 0 ; i < len; ++ i )
39          if ( value[i] > max )
40             max = value[i];
41
42     cout << max;
43
44      for ( int i = 0 ; i < len; ++ i )
45         value[i] = 1 ;
46
47      for ( int i = 0 ; i < len; ++ i )
48      {
49         max = 0 ;
50         t = 0 ;
51
52          for ( int j = 0 ; j < i; ++ j )
53              if ( a[i] > a[j] && value[j] > max )
54              {
55                 t = 1 ;
56                 max = value[j];
57             }
58
59          if ( t != 0 )
60             value[i] = max + 1 ;
61     }
62
63     max = 1 ;
64      for ( int i = 0 ; i < len; ++ i )
65          if ( value[i] > max )
66             max = value[i];
67
68     cout << ' ' << max;
69
70      return 0 ;
71 }
72

Darren 2008-08-19 16:32 发表评论