Bone Collector(hdoj--2602--01背包)
来源:互联网 发布:算法的时间复杂度定义 编辑:程序博客网 时间:2024/06/05 05:20
Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 40331 Accepted Submission(s): 16756
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
15 101 2 3 4 55 4 3 2 1
Sample Output
14
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct node{int v;int val;}edge[1010];int f[1010];int main(){int t;scanf("%d",&t);while(t--){memset(f,0,sizeof(f));memset(edge,0,sizeof(edge));int n,v;scanf("%d%d",&n,&v);for(int i=0;i<n;i++)scanf("%d",&edge[i].val);for(int i=0;i<n;i++)scanf("%d",&edge[i].v);for(int i=0;i<n;i++){for(int j=v;j>=edge[i].v;j--){f[j]=max(f[j],f[j-edge[i].v]+edge[i].val);}}printf("%d\n",f[v]);}return 0;}
0 0
- HDOJ 2602 Bone Collector (01背包)
- HDOJ 2602 Bone Collector(01背包)
- Bone Collector(hdoj--2602--01背包)
- HDOJ 2602 Bone Collector 01背包问题
- HDOJ 2602 Bone Collector (01背包)
- hdoj 2602 Bone Collector 【01背包】
- HDOJ 2602 Bone Collector【01背包】
- hdoj 2602 Bone Collector 【01-背包】
- HDOJ 2602 Bone Collector(01背包)
- hdoj 2602 Bone Collector 【01背包】
- HDOJ 2602 Bone Collector--01背包
- hdoj 2602 Bone Collector【01背包】
- HDOJ--2602--Bone Collector--dp--01背包
- HDOJ 2602 Bone Collector(背包问题)
- HDOJ 题目2602 Bone Collector(动态规划,01背包)
- HDOJ-----2602Bone Collector(DP-----01背包)
- HDOJ(HDU).2602 Bone Collector (DP 01背包)
- dp (0,1背包)Bone Collector HDOJ-2602 (基本01背包)
- 基于Spring注解的WEB MVC开发,URL映射
- linux 统计 程序运行时间
- 仿qq向左滑动删除 案例 详解
- 有了 Linux,你就可以搭建自己的超级计算机
- Unity3D里Time时间体系的讲解与运用
- Bone Collector(hdoj--2602--01背包)
- spring事物管理机制
- 项目管理心得:一个项目经理的个人体会、经验总结 [Y]
- ThreadLocal
- 【Ubuntu】Linux必备软件之Samba
- hdu 2546 饭卡(0-1背包)
- bzoj-1194 潘多拉的盒子
- Spring事务配置的五种方式
- 【POJ3468】【线段树成段更新】