Backpack problems, 0-1 backpack Problems

Backpack problems, 0-1 backpack Problems

Greedy

Description

Now there are a lot of items (they can be separated), and we know the value v and weight w of each item (1 <= v, w <= 10 ); if you have a backpack that can hold the weight of m (10 <= m <= 20), all you need to do is to pack the item into the backpack, this maximizes the total value of items in a backpack.

Input

Enter a positive integer n (1 <=n <= 5) In the first line, indicating that n groups of test data exist;
Then there is n test data. The first row of each group of test data has two positive integers s, m (1 <= s <= 10), and s indicates that there are s items. Each row in the next s line has two positive integers v, w.

Output

Outputs the value and value of the items in the backpack in each group of test data. Each output occupies one row.

Sample Input

13 155 102 83 9

Sample output

65

1 # include <stdio. h> 2 typedef struct value 3 {4 int v;/* value */5 int w;/* weight */6} thing; 7 int main () 8 {9 int n; 10 scanf ("% d", & n); 11 while (n --) 12 {13 int s, m, j, I = 0, sumv = 0, sumw = 0; 14 scanf ("% d", & s, & m); 15 thing a [s], t; 16 while (I! = S) 17 {18 scanf ("% d", & a [I]. v, & a [I]. w); 19 I ++; 20} 21 for (I = 0; I <s-1; I ++) 22 for (j = I; j <s; j ++) 23 {24 if (a [I]. v <a [j]. v) 25 {26 t = a [I]; 27 a [I] = a [j]; 28 a [j] = t; 29} 30} 31 int ww = 0, vv = 0; 32 I = 0; 33 while (sumw <m) 34 {35 ww = sumw; 36 vv = sumv; 37 sumw = sumw + a [I]. w; 38 sumv = sumv + a [I]. w * a [I]. v; 39 I ++; 40 while (I = s) 41 break; 42} 43 if (sumv> m) 44 sumv = vv + (m-ww) * a [I-1]. v; 45 printf ("% d \ n", sumv); 46} 47}