POJ 3624- Charm Bracelet（01背包 滚动数组）

Charm Bracelet

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 36395 Accepted: 15944

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 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

分析

01背包基础题

#include<iostream>
#include<string.h>
#include<algorithm>
#include<cstdio>
using namespace std;
const int maxv=112880+5;
const int maxn=3500;
int dp[maxv];
int c[maxn];
int w[maxn];
int main(){
int  n,V;
scanf("%d%d",&n,&V);
memset(dp,0,sizeof(dp));

for(int i=1;i<=n;i++)
scanf("%d%d",&c[i],&w[i]);

for(int i=1;i<=n;i++)
for(int j=V;j>=c[i];j--)
dp[j]=max(dp[j],dp[j-c[i]]+w[i]);

printf("%d\n",dp[V]);
return 0;
}