Nanyang 106--knapsack problem

Knapsack Problem time limit:MS | Memory limit:65535 KB Difficulty:3

Describe

Now there are a lot of items (they are divisible), we know the value of each unit weight of each item V and the weight w (1<=v,w<=10); If I give you a backpack, it can hold a weight of M (10<=m<=20), All you have to do is pack the items into your backpack so that the value of the items in your backpack is the largest.

Input

the first line enters a positive integer n (1<=n<=5), which indicates that there are n sets of test data;
Then there are n test data, the first line of each set of test data has two positive integers s,m (1<=s<=10); s indicates that there is an S item. The next S-line has two positive integer v,w per line.

Output

Outputs the value of the items in the

backpack for each set of test data and each output takes up one row.

Sample input

13 155 102) 83 9

Sample output

65

Source

[Miao Building] Original

Uploaded by

Miao-dong building

1#include <cstdio>2#include <algorithm>3 using namespacestd;4 5 structAC6 {7     intA, B;8} num[ the];9 Ten BOOLCMP (AC A, AC b) One { A     returnA.A >B.A; -  }  -   the intMain () - { -     intT; -scanf"%d", &t); +      while(t--) -     { +         intN, M; Ascanf"%d%d", &n, &m); at          for(intI=0; i<n; i++) -scanf"%d%d", &AMP;NUM[I].A, &num[i].b); -Sort (num, num+N, CMP); -         intsum =0, total =0; -          for(intI=0; i<m; i++) -         { insum + = num[i].b; Total + = num[i].a*num[i].b; -             //printf ("%d%d\n", sum, total); to             if(Sum >m) +             { -Total-= (sum-m) *num[i].a; the                  Break; *             } $             if(Sum = =m)Panax Notoginseng                  Break; -         } theprintf"%d\n", total); +     } A     return 0; the}

Nanyang 106--knapsack problem