【POJ】3624 - Charm Bracelet（01背包）

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Charm Bracelet

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 weightWi (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

很明显的01背包。

代码如下：

#include <cstdio>
#include <stack>
#include <queue>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define LL long long
struct node
{
	int w,v;
}a[3412];
int main()
{
	int n,m;
	int dp[13000];
	while (~scanf ("%d %d",&n,&m))
	{
		CLR(dp,0);		//初始化这样的错误好久没犯了，得长记性 
		for (int i = 1 ; i <= n ; i++)
			scanf ("%d %d",&a[i].w,&a[i].v);
		for (int i = 1 ; i <= n ; i++)
		{
			for (int j = m ; j >= a[i].w ; j--)
				dp[j] = max (dp[j] , dp[j-a[i].w] + a[i].v);
		}
		printf ("%d\n",dp[m]);
	}
	return 0;
}