动态规划进阶题目之Charm Bracelet

动态规划进阶题目之Charm Bracelet

时间: 1ms        内存:64M

描述:

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.

输入:

* 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

输出:

* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

示例输入:

4 6
1 4
2 6
3 12
2 7

示例输出:

23

提示:

参考答案(内存最优[1092]):

#include<stdio.h>
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};//赋零值,下面比较用
        for(i=1;i<=n;i++)
            scanf("%d%d",&w[i],&p[i]);
        for(i=1;i<=n;i++)//循环n件物品
            for(j=m;j>=w[i];j--)//循环重量,0到m,不断比较加上第i件物品与不加上第i件物品的价值,值大的赋值给c[j]
                c[j]=max(c[j],c[j-w[i]]+p[i]);
        printf("%d\n",c[m]);
    }
    return 0;
}

参考答案(时间最优[54]):

#include<stdio.h>

int n,max_w,max_d,i,t;
int w[3403],d[3403];
int f[12881];

int main()
{
    scanf("%d%d",&n,&max_w);
    for (i=1;i<=n;i++)
        scanf("%d%d",&w[i],&d[i]);
    for (i=1;i<=n;i++)
        for (t=max_w-w[i];t>=0;t--)
            if (f[t+w[i]]<f[t]+d[i])
                f[t+w[i]]=f[t]+d[i];
    for (t=0;t<=max_w;t++)
        if (f[t]>max_d)
            max_d=f[t];
    printf("%d\n",max_d);
    return 0;
} 

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