| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 45191 | Accepted: 19318 |
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
问题分析：
N 个物品每个物品有价值v[i]，重量w[i]， 给定背包最大承重M，求背包能够装载的最大价值。每个物品只有放入背包和不放入背包两种选择。
这是典型的0-1背包问题。
代码的时间上限是O(nm), 对于每个物品i, 它所要遍历的整数区间都是[ci, m]
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
using namespace std;
int dp[];
int w[];
int v[];
int main()
{
int n,m;
cin>>n>>m;
memset(dp,,sizeof(dp));
for(int i=;i<=n;i++)
{
cin>>w[i]>>v[i];
}
for(int i=;i<=n;i++)
{
for(int j=m;j>=w[i];j--)//要倒着推过来，因为dp【j】是滚动数组
{ dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
}
}
cout<<dp[m];
return ;
}