Charm Bracelet

题目链接：http://poj.org/problem?id=3624

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 47823		Accepted: 20322

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

Source

USACO 2007 December Silver

 

题目大意：输入n,m，n代表n个物品，m代表背包最多能背多重，下面n行每行两个数，分别代表每种物品的重量和价值，要求价值最大

思路：这题可以说是01背包的裸题了，但是这题不能用二维数组来做，会超内存，只能用一维

看代码：

#include<iostream>
#include<string.h>
#include<map>
#include<cstdio>
#include<cstring>
#include<stdio.h>
#include<cmath>
#include<ctype.h>
#include<math.h>
#include<algorithm>
#include<set>
#include<queue>
typedef long long ll;
using namespace std;
const ll mod=1e9;
const int maxn=3402+5;
const int maxm=12880+5;
const int maxx=1e4+10;
const ll maxe=1000+10;
#define INF 0x3f3f3f3f3f3f
#define Lson l,mid,rt<<1
#define Rson mid+1,r,rt<<1|1
struct p
{
    int w,v;
}a[maxn];
int dp[maxm];//dp[i]代表重量为i时最大价值
int main()
{
    int n,m;
    cin>>n>>m;
    memset(dp,0,sizeof(dp));
    for(int i=1;i<=n;i++)
    {
        cin>>a[i].w>>a[i].v;
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=m;j>=0;j--)//这里只能是从大往小遍历，因为如果从小往大了遍历的话，会重复用到一个物品
        {
            if(j>=a[i].w) dp[j]=max(dp[j],dp[j-a[i].w]+a[i].v);
        }
    }
    cout<<dp[m]<<endl;
    return 0;
}