题目

链接

http://poj.org/problem?id=3624

字面描述

Charm Bracelet
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 69960 Accepted: 28989
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

USACO 2007 December Silver

思路

01背包一维数组反向优化

代码实现

#include
#include
using namespace std;

const int maxn=5e3+10;
const int maxm=2e4+10;
int n,m;
struct node{
	int w,v;
	void input(){
		scanf("%d%d",&w,&v);
	}
}a[maxn];
int f[maxm];
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)a[i].input();
	for(int i=1;i<=n;i++){
		for(int j=m;j>=a[i].w;j--)f[j]=max(f[j],f[j-a[i].w]+a[i].v);
	}
	printf("%d",f[m]);
	return 0;
}
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