POJ(3624):Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 47057		Accepted: 20017

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

解题思路：

这是一道背包问题，要求在N件物品中挑选总体积不超过j的总价值最大的物品，每一件物品都有选和不选两种状态。对每一个物品依次遍历，求每一个容积选或不选该物品后的最大价值和。即为该容积在当前个数的物品条件下所能取得的价值最大总和。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn = 12880+20;
int W[maxn],D[maxn],dp[maxn];
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i = 1;i <= n;i++){
        scanf("%d%d",&W[i],&D[i]);
    }
    for(int k = 1;k <= n;k++){
        for(int w = m;w >= W[k];w--){
            dp[w] = max(dp[w],dp[w-W[k]] + D[k]);
        }
    }
    printf("%d",dp[m]);
}