POJ 3624 Charm Bracelet 01揹包問題

 Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15164		Accepted: 6936

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 iin 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

基礎揹包問題，大概意思是，有承重m的揹包；另有n個物品。第i個物品重w[i],價值p[i],求揹包能裝入物品的最大價值。
把重量設爲數組下標 即c[i],i爲物品重量，c[i]爲揹包重i時，物品的總價值。

一個公式：c[j]=max(c[j],c[j-w[i]]+p[i])

意思是，不加上第i個物品時揹包的價值，與重量j減去第i個物品重量後背包的價值加上第i個物品的價值比較，取較大者。

#include
int max(int x,int y)
{
    return x>y?x:y;
}
int main()
{
    int n,m,i,j;
    int w[4000],p[4000];
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        int c[20000]={0};//賦零值，下面比較用
        for(i=1;i<=n;i++)
            scanf("%d%d",&w[i],&p[i]);
        for(i=1;i<=n;i++)//循環n件物品
            for(j=m;j>=w[i];j--)//循環重量，m到w[i]，不斷比較加上第i件物品與不加上第i件物品的價值，值大的賦值給c[j]
                c[j]=max(c[j],c[j-w[i]]+p[i]);
        printf("%d\n",c[m]);
    }
    return 0;
}

參考