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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
#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;
#define Maxn 14000
int Val[Maxn];
int bag(int val[],int weight[],int M,int N)
{
memset(Val,0,sizeof(Val));
int maxn = 0;
for(int i = 1; i <= N; i++)
{
for(int j = M; j >= weight[i]; j--)
{
if (Val[j] < (Val[j - weight[i]] + val[i]))
{
Val[j] = max(Val[j],Val[j - weight[i]] + val[i]);
}
}
}
return Val[M];
}
int main()
{
int N,M;
int val[Maxn];
int weight[Maxn];
scanf("%d%d",&N,&M);
for(int i = 1; i <= N; i++)
{
scanf("%d",weight+i);
scanf("%d",val+i);
}
printf("%d\n",bag(val,weight,M,N));
return 0;
}
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原文地址：http://www.cnblogs.com/yakoazz/p/5718911.html
