Charm Bracelet
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 47080 | Accepted: 20026 |
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
1.就算用了动态规划，还是超时。因为状态转化方程dp[][]大小要开到12880，像如下代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[3600][13000];
int w[3600];
int money[3600];
int n,v;
int main()
{
cin>>n>>v;
for(int i=1;i<=n;i++)
{
scanf("%d%d",&w[i],&money[i]);
}
for(int j=1;j<=v;j++)
{
if(w[1]<=j)
{
dp[1][j]=money[j];
}
else
{
dp[1][j]=0;
}
}
for(int i=1;i<=n;i++)
{
for(int j=1;j<=v;j++)
{
if(j-w[i]>=0)
dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+money[i]);
else
dp[i][j]=dp[i-1][j];
}
}
cout<<dp[n][v]<<endl;
return 0;
}2.因为它在更新的时候，只与上一行有关（如下面表格：详细背包讲解，见点击转到）可以考虑用滚动数组存储。因为是从左往右更新，用滚动数组的时候可以从右往左覆盖。
表格：
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int w[3600];
int money[3600];
int dp[13000];
int n,v;
int main()
{
cin>>n>>v;
for(int i=0;i<n;i++)
scanf("%d%d",&w[i],&money[i]);
memset(dp,0,sizeof(dp));
for(int i=0;i<n;i++)
{
for(int j=v;j>=w[i];j--)
dp[j]=max(dp[j-w[i]]+money[i],dp[j]);
}
cout<<dp[v]<<endl;
return 0;
}