最经典、最基础的01背包入门题目
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2602
Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13839 Accepted Submission(s): 5454
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 ?
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.
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
Author
Teddy
Source
Recommend
lcy
很基础的背包入门题目，如果看了代码还不会做，就自己去百度文档中下载《背包九讲》吧。很强大的，注意有个事不要积分就可以下载的，看清楚了啊！
给几个下载地址：http://wenku.baidu.com/view/16527702de80d4d8d15a4f06.html
http://wenku.baidu.com/view/16527702de80d4d8d15a4f06.html
//hdu 2602 AC 0ms 248k
#include<stdio.h>
#include<string.h>
int main()
{
int test;
int n,v;
int i,j;
int c[1005],w[1005],sum[1005];
scanf("%d",&test);
while(test--)
{
memset(sum,0,sizeof(sum));
scanf("%d%d",&n,&v);
for(i=0;i<n;i++)
scanf("%d",&w[i]);//输入价值
for(i=0;i<n;i++)
scanf("%d",&c[i]);//输入对应的体积
for(i=0;i<n;i++)
for(j=v;j>=c[i];j--)
{
if(sum[j]<=(sum[j-c[i]]+w[i]))
sum[j]=sum[j-c[i]]+w[i];
}
printf("%d\n",sum[v]);
}
return 0;
}
下面再贴个另外的代码，千万别提交啊，不是这道题目的。
/*
//注意下面的不要提交啊~~~
//不过貌似是另外一道题目的AC代码，具体是哪道题目我忘了。。。
//看了kb神的另外一篇完全背包博客后从新写的，如果改变测试数据输入的顺序就可如此AC，应该可以省下不少内存，虽然内存不要钱，但是能省则省了哈！O(∩_∩)O哈哈哈~大神果然强大
#include<stdio.h>
#include<string.h>
int main()
{
int test;
int n,v;
int i,j;
int c,w,sum[1005];
scanf("%d",&test);
while(test--)
{
memset(sum,0,sizeof(sum));
scanf("%d%d",&n,&v);
while(n--)
{
scanf("%d%d",&w,&c);
for(i=v;i>=c;i--)
if(sum[i]<=(sum[i-c]+w))
sum[i]=sum[i-c]+w;
}
printf("%d\n",sum[v]);
}
return 0;
}
*/