POJ 3624 Charm Bracelet （01背包）

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 33959		Accepted: 15060

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

Source

USACO 2007 December Silver

简单的01背包模型

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
int w,d;
}t[110000];
int dp[110000];
int main()
{
int n,m,i,j,k,l;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(i=0;i<n;i++)
scanf("%d%d",&t[i].w,&t[i].d);
for(i=0;i<n;i++)
for(j=m;j>=t[i].w;j--)
{
dp[j]=max(dp[j],dp[j-t[i].w]+t[i].d);
}
printf("%d\n",dp[m]);
}
return 0;
}