POJ 3624 Charm Bracelet（DP 01背包）

Charm Bracelet Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 33482   Accepted: 14839 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 weightWi (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背包问题，入门题目。

下面是AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;

struct node
{
int h,val;
}a[35000];
int dp[350000];

int main()
{
int n,v;
while(~scanf("%d%d",&n,&v))
{
for(int i=1;i<=n;i++)
{
cin>>a[i].h>>a[i].val;
}
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
{
for(int j=v;j>=a[i].h;j--)
{
dp[j]=max(dp[j],dp[j-a[i].h]+a[i].val);
}
}
printf("%d\n",dp[v]);
}
return 0;
}