POJ 3624 Charm Bracelet

Charm Bracelet

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 20499 Accepted: 9293

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 weightW_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

Source

USACO 2007 December Silver

题解：经典背包问题，基础01背包，状态转移方程：

f[v]=max(f[v],f[v-a[i].w]+a[i].d);

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int n,m;int f[15000];struct thing{  int w;  int d;}a[5000];int main(){    //freopen("in.txt","r",stdin);    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++)    {      scanf("%d%d",&a[i].w,&a[i].d);      f[i]=0;    }    for(int i=1;i<=n;i++)    {      for(int v=m;v>=a[i].w;v--)      {        f[v]=max(f[v],f[v-a[i].w]+a[i].d);      }    }    //for(int i=1;i<=m;i++)printf("%d\n",f[i]);    printf("%d\n",f[m]);    return 0;}