Charm Bracelet

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

Sample Output

C++版本一

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>using namespace std;
int n,m;
struct node{int d,w;}b[4000];
bool cmp(node a,node b){if(a.w==b.w)return a.d>a.d;return a.w>b.w;}
int dp[13005];
int main()
{scanf("%d%d",&n,&m);for(int i=0;i<n;i++){scanf("%d%d",&b[i].w,&b[i].d);}memset(dp,0,sizeof(dp));sort(b,b+n,cmp);for(int i=0;i<n;i++){//dp[i]=b[i].d;for(int j=m;j>=b[i].w;j--){dp[j]=max(dp[j],dp[j-b[i].w]+b[i].d);}}cout << dp[m] << endl;//cout << "Hello world!" << endl;return 0;
}

C++版本二

///2014.4.10
///poj3624#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;int N,M;
int w[3500],c[3500];
int f[13500];int main()
{// freopen("in","r",stdin);// freopen("out","w",stdout);cin>>N>>M;for(int i=1 ; i<=N ; i++){cin>>c[i]>>w[i];}memset(f,0,sizeof(f) );for(int i=1 ; i<=N ; i++){for(int j=M ; j>=1 ; j--){int a;if( j-c[i]>=0 )a = f[j-c[i]]+w[i];elsea = 0;f[j] = f[j]>a? f[j]:a;}}cout<<f[M]<<endl;return 0;
}

C++版本三

#include <iostream>
using namespace std;  const int MAX_N = 3405;
const int MAX_M = 12885;  int dp[MAX_M];
int w[MAX_N];
int d[MAX_N];  int max(int a, int b) {  return a > b ? a : b;
}  int main()
{  int n, m;  cin >> n >> m;  for(int i = 1; i <= n; i++) {  cin >> w >> d;  }  for(int i = 1; i <= n; i++) {  for(int j = m; j >= w; j--) {  dp[j] = max(dp[j-w] + d, dp[j]);  }  }  cout << dp[m] << endl;  return 0;
}

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