poj3624 Charm Bracelet

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Charm Bracelet

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 22920 Accepted: 10315

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

背包问题

源代码：

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>#include<math.h>#include<cstdlib>using namespace std;#include<stdio.h>#define N 300000+10int w[N],d[N],f[N];int main(){    int n , m, i , j;    while(scanf("%d%d",&n,&m)!=EOF)    {        memset(f,0,sizeof(f));        for(i=1; i<=n; i++)            scanf("%d%d",&w[i],&d[i]);        for(i=1; i<=n; i++)            for(j=m; j>=w[i]; j--)                f[j]=max(f[j],f[j-w[i]]+d[i]);        printf("%d\n",f[m]);    }    return 0;}