poj 3624(01背包)

Charm Bracelet

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 31347 Accepted: 13942

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

Source

USACO 2007 December Silver

//裸的01背包

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;int w[4000],v[4000];int dp[12885];int n,m;void ZeroOnePack(int w,int v){    for(int i=m;i>=w;i--)        dp[i]=max(dp[i],dp[i-w]+v);}int main(){    while(~scanf("%d%d",&n,&m))    {        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;i++)            scanf("%d%d",w+i,v+i);        for(int i=1;i<=n;i++)            ZeroOnePack(w[i],v[i]);        printf("%d\n",dp[m]);    }    return 0;}