HDU 2062 Bone Collector

来源：互联网发布：怎样装修淘宝店铺首页编辑：程序博客网时间：2024/05/19 14:35

Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 67578 Accepted Submission(s): 28218

Problem Description
Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?

Input
The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

Output
One integer per line representing the maximum of the total value (this number will be less than 231).

Sample Input

1
5 10
1 2 3 4 5
5 4 3 2 1

Sample Output

14

代码

#include<stdio.h>
#include<string.h>
int main()
{
    int i,j,n,v,m,m1,x,p[1005],t,f[1005],w[1005];
    scanf("%d",&t);
    while(t--){
        memset(f,0,sizeof(f));
        scanf("%d %d",&n,&v);
        for(i=0;i<n;i++)
            scanf("%d",&p[i]);
        for(i=0;i<n;i++)
            scanf("%d",&w[i]);
        for(i=0;i<n;i++)
        for(j=v;j>=w[i];j--){
            f[j]=f[j]>(f[j-w[i]]+(p[i]))?f[j]:(f[j-w[i]]+(p[i]));
        }
        printf("%d\n",f[v]);
    }
    return 0;
}

每次做完题再看别人代码总有新发现，别人价值和体积是用一个结构体写的，好高大上，下一次我也用

阅读全文

0 0