hdu——2602Bone Collector(第一类背包问题)
2016-04-19 20:41
288 查看
Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 46744 Accepted Submission(s): 19463
[align=left]Problem Description[/align]
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 ?
[align=left]Input[/align]
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.
[align=left]Output[/align]
One integer per line representing the maximum of the total value (this number will be less than 231).
[align=left]Sample Input[/align]
1 5 10 1 2 3 4 5 5 4 3 2 1
[align=left]Sample Output[/align]
14
[align=left]Author[/align]
Teddy
[align=left]Source[/align]
HDU 1st “Vegetable-Birds Cup” Programming Open Contest
[align=left]Recommend[/align]
lcy | We have carefully selected several similar problems for you: 1203 2159 1171 2191 2844
第一类背包 for(i=1;i<=n;i++)
for(j=V;j>v[i];j--)
d[j]=max(d[j],d[j-v[i]]+jg[i])
#include<iostream> #include<cstdio> #include<cstring> #include<string> #include<queue> #include<set> #include<map> #include<sstream> #include<algorithm> #include<cmath> #include<cstdlib> #include<deque> using namespace std; struct bb { long long v; long long jz; }a[1005]; long long max(long long n,long long m) { if(n>=m)return n; else return m; } int main() { long long n,m,vv,sum[1005]; cin>>n; while(n--) { cin>>m>>vv; memset(sum, 0, sizeof(sum)); for(int i=0;i<m;i++) { cin>>a[i].jz; } for(int i=0;i<m;i++) { cin>>a[i].v; } for(int i=0;i<m;i++) { for(int j=vv;j>=a[i].v;j--) { sum[j]=max(sum[j],sum[j-a[i].v]+a[i].jz); } } cout<<sum[vv]<<endl;//注意是vv而不是其他 } return 0; }
相关文章推荐
- 【java项目实战】dom4j解析xml文件,连接Oracle数据库
- 构建一个更好的人脸追踪器---31
- 软考信息系统监理师,2016年4月15日作业
- hdu1068(二分)Girls and Boys
- 山东省第五届ACM大学生程序设计竞赛 Full Binary Tree
- 关于颜色色值和对应名称的总结
- 团队作业3
- Mybatis最入门---ResultMaps基本用法
- 用C语言实现MYSQL的简单操作
- 关于颜色色值和对应名称的总结
- 【java项目实战】Servlet详解以及Servlet编写登陆页面(二)
- 正则表达式
- swift开发笔记24 解决键盘遮挡输入框 的方法
- UVA 10192 Vacation
- (转载)动态规划之背包问题(一)
- [置顶] JAVA 压缩图片-解析一维码二维码
- Vbs备份数据脚本集合
- ZOJ 3600 Taxi Fare (细节题)
- JVM内存回收机制简述
- 双向循环链表的头插、中插、尾插、删除、逆序顺序显示(C++实现)