NYOJ-103-backpack Problems

Question address: http://acm.nyist.net/JudgeOnline/problem.php? PID = 1, 106

 

Backpack problem time limit: 3000 MS | memory limit: 65535 kb difficulty: 3

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

1 3 15 5 10 2 8 3 9 

Sample output

65

 

Solution report: the most basic knapsack problem, starting from the most valuable, can be loaded, full. We recommend that you read the nine articles about backpacks.

 

1

2 # include <stdio. h>
3 # include <string. h>
4 # include <stdlib. h>
5
6 struct P {
7 int V;
8 int W;
9} p [15];
10
11 int CMP (const void * a, const void * B ){
12 struct p * c = (struct p *);
13 struct p * D = (struct p *) B;
14 return D-> V-C-> V;
15}
16
17 int main (){
18 int N, S, M, A, B, I, ans;
19 scanf ("% d", & N );
20 while (n --){
21 scanf ("% d", & S, & M );
22 For (I = 0; I <s; I ++ ){
23 scanf ("% d", & P [I]. V, & P [I]. W );
24}
25 qsort (P, S, sizeof (P [0]), CMP );
26 I = ans = 0;
27 while (M> = P [I]. W ){
28 ans + = P [I]. W * P [I]. V;
29 m-= P [I]. W;
30 I ++;
31}
32 ans + = m * P [I]. V;
33 printf ("% d \ n", ANS );
34}
35 return 0;
36}
37

NYOJ-103-backpack Problems