POJ - 3624
Description Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight Wi (1 ≤ Wi ≤ 400), a 'desirability' factor Di (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880). Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings. Input * Line 1: Two space-separated integers: N and M Output * Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints Sample Input 4 6 1 4 2 6 3 12 2 7 Sample Output 23 Source
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[100100];
struct node
{
int w,val;
}edge[100100];
int main()
{
int m,n;
while(scanf("%d%d",&m,&n)!=EOF)
{
memset(dp,0,sizeof(dp));
for(int i=0;i<m;i++)
scanf("%d%d",&edge[i].w,&edge[i].val);
for(int i=0;i<m;i++)
for(int j=n;j>=edge[i].w;j--)
dp[j]=max(dp[j],dp[j-edge[i].w]+edge[i].val);
printf("%d
",dp[n]);
}
return 0;
} |
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