poj 3624 Charm Bracelet (01背包 基础)

Charm Bracelet

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 34504 Accepted: 15285

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 61 42 63 122 7

Sample Output

23

一道01背包的基础题，这题用的是优化后的一维数组存放最大价值:

f[j]=max(f[j],f[j-w[i]]+c[i]])

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int f[14000];int main(){    int n,m;    int w[4000],c[4000];    while(~scanf("%d%d",&n,&m))    {        memset(f,0,sizeof(f));        for(int i=1;i<=n;i++)        {            scanf("%d%d",&w[i],&c[i]);        }        for(int i=1;i<=n;i++)        {            for(int j=m;j>=w[i];j--)                f[j]=max(f[j],f[j-w[i]]+c[i]);        }        printf("%d\n",f[m]);    }    return 0;}