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  • HDU 3732 Ahui Writes Word

      

      

    Ahui Writes Word

    Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 434 Accepted Submission(s): 182


    Problem Description

    We all know that English is very important, so Ahui strive for this in order to learn more English words. To know that word has its value and complexity of writing (the length of each word does not exceed 10 by only lowercase letters), Ahui wrote the complexity of the total is less than or equal to C.
    Question: the maximum value Ahui can get.
    Note: input words will not be the same.

    Input

    The first line of each test case are two integer N , C, representing the number of Ahui’s words and the total complexity of written words. (1 ≤ N ≤ 100000, 1 ≤ C ≤ 10000)
    Each of the next N line are a string and two integer, representing the word, the value(Vi ) and the complexity(Ci ). (0 ≤ Vi , Ci ≤ 10)

    Output

    Output the maximum value in a single line for each test case.

    Sample Input

    5 20 go 5 8 think 3 7 big 7 4 read 2 6 write 3 5

    Sample Output

    15
    Hint
    Input data is huge,please use “scanf(“%s”,s)”

    Author

    Ahui

    Source

    Recommend

    notonlysuccess
     
     1 #include<stdio.h>
    2 #include<string.h>
    3 char ch[100] ;
    4 int nkind , total_comp ;
    5 int value[100024] , comp[100024] , fine[100024] , map[11][11];
    6 int main ()
    7 {
    8 int pos , x , y ,sum , t ;
    9 while ( scanf ( "%d%d" , &nkind , &total_comp ) != EOF )
    10 {
    11 memset( map , 0 , sizeof (map) ) ;
    12 for ( int i = 0 ; i < nkind ; i ++ )
    13 {
    14 scanf ( "%s%d%d" , ch , &x , &y ) ;
    15 map[x][y] ++ ;
    16 }
    17 pos = 0 ; // 用二分制转换成01背包
    18 for ( int i = 0 ; i <= 10 ; i ++ )
    19 for ( int j = 0 ; j <= 10 ; j ++ )
    20 {
    21 if( map[i][j] >= 1 )
    22 {
    23 sum = 1 , t = 1 ;
    24 while ( sum <= map[i][j] )
    25 {
    26 value[pos] = i * t ;
    27 comp[pos++] = j * t ;
    28 t *= 2 ;
    29 sum += t ;
    30 }
    31 if ( ( sum - t ) < map[i][j] )
    32 {
    33 value[pos] = i * (map[i][j]-sum + t) ;
    34 comp[pos++] = j * (map[i][j]-sum + t) ;
    35 }
    36 }
    37 }
    38 memset( fine , 0 , sizeof (fine) ) ;
    39 for ( int i = 0 ; i < pos ; i ++ )
    40 for ( int j = total_comp ; j >= comp[i] ; -- j )
    41 if ( fine[j] < fine[j-comp[i]] + value[i] )
    42 fine[j] = fine[j-comp[i]] + value[i] ;
    43 printf ( "%d\n" , fine[total_comp] ) ;
    44 }
    45 return 0 ;
    46 }

      

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  • 原文地址:https://www.cnblogs.com/jbelial/p/2116160.html
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