hdu 2602 Bone Collector (01背包经典入门)
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 ?
![](http://acm.hdu.edu.cn/data/images/C154-1003-1.jpg)
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 ?
![](http://acm.hdu.edu.cn/data/images/C154-1003-1.jpg)
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; }