标签：des style blog color os 2014
<span style="color:#3333ff;">/* __________________________________________________________________________________________________ * copyright: Grant Yuan * * algorithm: 01背包（就地滚动） * * time : 2014.7.18 * * declare : 题目中说N最大是3400多，但是一开始开了5000内存还是运行时错误，后来直接改了50000 * * * *_________________________________________________________________________________________________*</span>
<span style="color:#3333ff;">
I - 01背包（就地滚动）
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit
Status
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
*/
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int w[50001];
int p[50001];
int sum;
int n;
int dp[50001];
int main()
{
cin>>n>>sum;
for(int i=0;i<n;i++)
cin>>w[i]>>p[i];
memset(dp,0,sizeof(dp));
for(int i=0;i<n;i++)
for(int j=sum;j>=w[i];j--)
{
dp[j]=max(dp[j],dp[j-w[i]]+p[i]);
}
cout<<dp[sum]<<endl;
return 0;
}
</span>标签：des style blog color os 2014
原文地址：http://blog.csdn.net/yuanchang_best/article/details/37933477
