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【POJ - 3624】Charm Bracelet

热度:19   发布时间:2023-12-06 19:46:05.0

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 Wi (1 ≤ Wi ≤ 400), a 'desirability' factor Di (1 ≤ Di ≤ 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: 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 6
1 4
2 6
3 12
2 7

Sample Output

23

解题报告:

又水了一道题目,这是一道裸的01背包问题,可是我又把数组开小了。。。。

ac代码:

#include<cstdio>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring> 
#include<queue>
#include<stack>
using namespace std;
int w[3500],val[3500];
int dp[20000];
int main()
{int n,m;scanf("%d%d",&n,&m);for(int i=0;i<n;i++)scanf("%d%d",&w[i],&val[i]);for(int i=0;i<n;i++){for(int j=m;j>=w[i];j--){dp[j]=max(dp[j],dp[j-w[i]]+val[i]);}}printf("%d\n",dp[m]);return 0;
}