Charm Bracelet
Time Limit: 1000MS | Memory Limit: 65536K |
Total Submissions: 7247 | Accepted: 3278 |
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
题意:给定每个物品的重量和价值,求给定最大重量下能获得的最大价值。。
我的背包第一题
思路:01背包,状态方程为dp[i][j]=MAX(dp[i-1][j],dp[i-1][j-weight[i]]+value[i])
#include<iostream>
using namespace std;
int N,W,weight[40000],cost[40000],dp[40000],i,j;
int main()
{
while(cin>>N>>W&&N)
{
for(i=0;i<N;i++)
cin>>cost[i]>>weight[i];
memset(dp,0,sizeof(dp));
for(i=0;i<N;i++)
for(j=W;j>=cost[i];j--)
{
if(j-cost[i]>=0)
{
if(dp[j-cost[i]]+weight[i]>dp[j])
dp[j]=dp[j-cost[i]]+weight[i];
}
}
cout<<dp[W]<<endl;
}
}