| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 23359 | Accepted: 10532 |
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
* Lines 2..N+1: Line i+1 describes charm i with two space-separated integers:
Wi and Di
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
还是01背包
#include<iostream>
#include<cstdio>
#include<cmath>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
int main()
{
int n,i,j,v;
while(scanf("%d%d",&n,&v)!=EOF){
int dp[13000]={0},c[3500],w[3500];
for(i=0;i<n;i++)
scanf("%d%d",c+i,w+i);
for(i=0;i<n;i++)
{
for(j=v;j>=c[i];j--)
{
if(dp[j]<dp[j-c[i]]+w[i])
{
dp[j]=dp[j-c[i]]+w[i];
}
}
}
printf("%d\n",dp[v]);
}
return 0;
}
POJ3624Charm Bracelet（01背包）,布布扣,bubuko.com
原文地址：http://blog.csdn.net/fljssj/article/details/38486651
