Charm Bracelet POJ - 3624

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 W i (1 ≤ W i ≤ 400), a ‘desirability’ factor D i (1 ≤ D i ≤ 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: W i and D i

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

思路：

赤裸裸的01背包，背就完了

Sample Output

23

思路：

赤裸裸的01背包，背就完了

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<iomanip>
#include<cstring>
#include<string>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define ll long long
#define mes(x,y); memset(x,y,sizeof(x))
#define mv 2147483648+30
using namespace std;
ll gar(ll a,ll b){//最大公约数 
return b==0?a:gar(b,a%b);
} 
struct node{
	int x,y;
}a[21000];
int dp[21000];
int main(){
	int n,m;
	while(cin>>n>>m){
		mes(a,0);mes(dp,0);
		for(int i=1;i<=n;i++){
			cin>>a[i].x>>a[i].y;
		}
		for(int i=1;i<=n;i++){
			for(int j=m;j>=a[i].x;j--){
				dp[j]=max(dp[j],dp[j-a[i].x]+a[i].y);
			}
		}
		cout<<dp[m]<<endl;
	}
}