nyoj~106~ knapsack problem ~ greedy algorithm ~

Knapsack Problem time limit: theMs | Memory Limit:65535KB 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 you give you a backpack, it can hold a weight of M (10<=m<=20), all you have to do is pack the items in your backpack. , so that the total value of the items in the 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 has n test data, and the first row of each set of test data has two positive integers s,m (1&l T;=S<=10); s indicates that there is an S item. The next S-line has two positive integer v,w per line. The

output

outputs the value of the items in the backpack in each set of test data, and each output takes up one row.

Sample Input

13 155 102 9 

sample output

65 

 ac-code:  

 
 #include <stdio.h> #include <algorithm>using namespace std;struct Bag{int v;int W;}; BOOL CMP (bag A,bag b) {return A.V>B.V;} int main () {int sum,s,m,i,max,n;bag a[100];scanf ("%d", &n), while (n--) {scanf ("%d%d", &s,&m); for (i=0;i< s;i++) scanf ("%d%d", &A[I].V,&A[I].W); Max=m;sum=0;sort (a,a+s,cmp); for (i=0;i<s;i++) {if (A[i].w<=max) {SUM+=A[I].V*A[I].W;MAX-=A[I].W;} Else{sum+=a[i].v*max;break;}} printf ("%d\n", sum);} return 0;} 
 
 

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nyoj~106~ knapsack problem ~ greedy algorithm ~