Charm Bracelet （01背包）

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 W_i (1 ≤ W_i ≤ 400), a 'desirability' factor D_i (1 ≤ D_i ≤ 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: 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
题目大意：这是一道裸01背包问题，没什么花样，可以当模板题来看。        Bessie去商场买珠宝，她要往手链上镶嵌珠宝，但是有重量限制，每块珠宝有两个属性，重量，价值（珠宝不可切割），问在重量不超额的情况下，价值最高多少。

01背包 属于是贪心中的一个分支，个人认为01背包关键在于状态的保存和更新，即dp【i】的更新。

01背包 01就是指该件物品仅有一件舍或是取（0 or 1）属于背包问题中比较简单的一类，就01背包来讲，有两种方法可解，一维背包和二维背包。


一维状态转移方程（val价值，weight重量，w最大承重，n物品件数）

    for（int i=1;i<=n;i++）
        for(int j=v;j>=weight[i];j--)//这里是关键点，一定要逆序
            dp[j]=max(dp[j],dp[j-wegiht[i]]+val[i]);

二维状态转移方程（变量同一维）

  dp[i][v]= max(dp[i-1][v], dp[i-1][v-weight[i]]+val[i]); 

就这道题，我是用一维过的

代码贴出来

#include <iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<ctype.h>
#include<stdlib.h>
#include<vector>
#include<queue>
using namespace std;
#define oo 0x3f3f3f3f
#define maxn 20010

int dp[maxn],weight[maxn],val[maxn];

void Init()
{
    memset(dp,0,sizeof(dp));
    memset(val,0,sizeof(val));
    memset(weight,0,sizeof(weight));
}

int main()
{
    int n,w;
    while(scanf("%d%d",&n,&w)!=EOF)
    {
        Init();
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&weight[i],&val[i]);
        }
        for(int i=1;i<=n;i++)
            for(int j=w;j>=weight[i];j--)
        {
            dp[j]=max(dp[j],dp[j-weight[i]]+val[i]);
        }
        printf("%d
",dp[w]);
    }
    return 0;
}