POJ3624Charm Bracelet（01背包）

標籤：01背包   acm   poj   

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23359		Accepted: 10532

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 theN (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weightWi (1 ≤ Wi ≤ 400), a ‘desirability‘ factorDi (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more thanM (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:Wi and Di

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

還是01背包

#include<iostream>#include<cstdio>#include<cmath>#include<queue>#include<algorithm>#include<cstring>using namespace std;int main(){    int n,i,j,v;    while(scanf("%d%d",&n,&v)!=EOF){        int dp[13000]={0},c[3500],w[3500];        for(i=0;i<n;i++)            scanf("%d%d",c+i,w+i);        for(i=0;i<n;i++)        {            for(j=v;j>=c[i];j--)            {                if(dp[j]<dp[j-c[i]]+w[i])                {                    dp[j]=dp[j-c[i]]+w[i];                }            }        }        printf("%d\n",dp[v]);    }    return 0;}