poj3624(01背包)

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17288		Accepted: 7850

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 theN (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weightW_i (1 ≤ W_i ≤ 400), a 'desirability' factorD_i (1 ≤ D_i ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more thanM (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:W_i and D_i

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

 

本题要求在给定的载重范围内是选择物品的价值最大。

本题每件物品只有一个，是个典型的01背包问题，传统做法开个二维数组dp[i][j],载重量为j的情况下考虑到了前i个物品，但本题可能超内存；开个一维数组，应该逆序避免重复。

#include<iostream>
#include<cstdio>
using namespace std;

struct bag
{
	int w,d;
}Bag[40000];
int f[12000+1000];

int Max(int a,int b)
{
	return a<b?b:a;
}

int main()
{
	int i,j,n,m;
	while(~scanf("%d%d",&n,&m))
	{
		for(i=1;i<=n;i++)
		{
			scanf("%d%d",&Bag[i].w,&Bag[i].d);
		}
		
		memset(f,0,sizeof(f));
		for(i=1;i<=n;i++)
		{
			for(j=m;j>=Bag[i].w;j--)
			{
				f[j]=Max(f[j],f[j-Bag[i].w]+Bag[i].d);
			}
		}

		printf("%d\n",f[m]);
	}
	return 0;
}