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): 77450    Accepted Submission(s): 32095

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
 
Author
Teddy
 
Source
 
分析:
经典的01背包问题,每个物品只有两种状态,放还是不放
dp[i][j]:即前面i件物品放入一个容量为j的背包可以获得的最大价值
状态转移方程:
dp[i][j]=dp[i-1][j]  j<w[i]
dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+v[i]     j>=w[i]
 
注意:
1.先输入的是物品的价值,而不是重量,题目中好像说反了
2.背包容量可以为0,物品的重量也可以为0
关于第2点的测试数据:
输入
1
5 0
2 4 1 5 1
0 0 1 0 0
输出
12
 
从前往后遍历二维数组
#include<bits/stdc++.h>
using namespace std;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,c;
scanf("%d %d",&n,&c);
int v[n+],w[n+];
for(int i=;i<=n;i++)
{
scanf("%d",&v[i]);
}
for(int i=;i<=n;i++)
{
scanf("%d",&w[i]);
}
int dp[n+][c+];
memset(dp,,sizeof(dp));
for(int i=;i<=n;i++)
{
for(int j=;j<=c;j++)
{
if(w[i]<=j)//表示第i个物品放入背包中
{
dp[i][j]=max(dp[i-][j],dp[i-][j-w[i]]+v[i]);//第i个物品放入之后,那么前面i-1个物品可能会因为剩余空间不够无法放入
}else//表示第i个物品不放入背包
{
dp[i][j]=dp[i-][j];//如果第i个物品不放入背包,那么此时的最大价值与放前面i-1个物品的值相等
}
}
}
printf("%d\n",dp[n][c]);
}
return ;
} /*
3
5 10
6 3 5 4 6
2 2 6 5 4
5 10
1 2 3 4 5
5 4 3 2 1
5 0
2 4 1 5 1
0 0 1 0 0 15
14
12
*/
 从后往前遍历二维数组
#include<bits/stdc++.h>
using namespace std;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,c;
scanf("%d %d",&n,&c);
int v[n+],w[n+];
for(int i=;i<=n;i++)
{
scanf("%d",&v[i]);
}
for(int i=;i<=n;i++)
{
scanf("%d",&w[i]);
}
int dp[n+][c+];
memset(dp,,sizeof(dp));
for(int j=;j<=c;j++)
{
if(j>=w[n])
{
dp[n][j]=v[n];
}else
{
dp[n][j]=;
}
}
for(int i=n-;i>=;i--)
{
for(int j=;j<=c;j++)
{
if(j>=w[i])
{
dp[i][j]=max(dp[i+][j],dp[i+][j-w[i]]+v[i]);
}
else
{
dp[i][j]=dp[i+][j];
}
}
}
printf("%d\n",dp[][c]);
}
return ;
} /*
3
5 10
6 3 5 4 6
2 2 6 5 4
5 10
1 2 3 4 5
5 4 3 2 1
5 0
2 4 1 5 1
0 0 1 0 0 15
14
12
*/

二者其实是一个方法,只是遍历二维数组的方向有点不同

HDU 2602 Bone Collector(经典01背包问题)的更多相关文章

  1. HDOJ(HDU).2602 Bone Collector (DP 01背包)

    HDOJ(HDU).2602 Bone Collector (DP 01背包) 题意分析 01背包的裸题 #include <iostream> #include <cstdio&g ...

  2. hdu 2602 Bone Collector(01背包)模板

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2602 Bone Collector Time Limit: 2000/1000 MS (Java/Ot ...

  3. 题解报告:hdu 2602 Bone Collector(01背包)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2602 Problem Description Many years ago , in Teddy’s ...

  4. hdu 2602 - Bone Collector(01背包)解题报告

    Bone Collector Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) T ...

  5. hdu 2602 Bone Collector(01背包)

    题意:给出包裹的大小v,然后给出n块骨头的价值value和体积volume,求出一路下来包裹可以携带骨头最大价值 思路:01背包 1.二维数组(不常用 #include<iostream> ...

  6. HDU - 2602 Bone Collector(01背包讲解)

    题意:01背包:有N件物品和一个容量为V的背包.每种物品均只有一件.第i件物品的费用是volume[i],价值是value[i],求解将哪些物品装入背包可使价值总和最大. 分析: 1.构造二维数组: ...

  7. HDU 2602 Bone Collector 0/1背包

    题目链接:pid=2602">HDU 2602 Bone Collector Bone Collector Time Limit: 2000/1000 MS (Java/Others) ...

  8. HDU 2602 Bone Collector

    http://acm.hdu.edu.cn/showproblem.php?pid=2602 Bone Collector Time Limit: 2000/1000 MS (Java/Others) ...

  9. HDU 2602 Bone Collector (01背包问题)

    原题代号:HDU 2602 原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=2602 原题描述: Problem Description Many yea ...

随机推荐

  1. hello Groovy

    Groovy [rocky@www ~]$ curl -s get.sdkman.io 1. 下载 [rocky@www Downloads]$ wget https://dl.bintray.com ...

  2. Django中间件解析

    一,中间件的概念 django 中的中间件(middleware),在django中,中间件其实就是一个类,在请求到来和结束后,django会根据自己的规则在合适的时机执行中间件中相应的方法.在dja ...

  3. HTML利用posotion属性定位 小技巧

    1.居中效果 父级DIV (index-top )属性设置为 text-align:center; 子级DIV( tabIndex-main)属性设置为 margin:0 auto;   2.左右对齐 ...

  4. Object equals 方法

    package com.mydemo.controller; public class TestEquals { public static void main(String[] args) { Do ...

  5. Android组件系列----当前Activity跳转到另一个Activity的详细过程

    [声明] 欢迎转载,但请保留文章原始出处→_→ 生命壹号:http://www.cnblogs.com/smyhvae/ 文章来源:http://www.cnblogs.com/smyhvae/p/3 ...

  6. ASP.NET中使用UpdatePanel时用Response输出出现错误的解决方法

    asp.net中执行到Response.write("xx");之类语句或Microsoft JScript 运行时错误: Sys.WebForms.PageRequestMana ...

  7. Vue2学习笔记:v-for指令

    1.使用 <!DOCTYPE html> <html> <head> <title></title> <meta charset=&q ...

  8. [TSQL|SQLSERVER|MSSQL数据库] 将数据库文件与日志附加到数据库引擎,以及转移数据库文件位置

    附加: USE [master] GO CREATE DATABASE [database_name] ON ( FILENAME = N'C:\Data\<database name>. ...

  9. C#实现ADH815通讯

    最近在做自提柜项目,考虑到ADH815电路板在自助售卖行业的通用性.把通讯代码贴出来了. 下载地址

  10. Oracle GoldenGate DDL 详细说明 使用手册(较早资料)

    一. 概述 DDL 相关的参数包括:DDL.DDLERROR.DDLOPTIONS.DDLSUBST.DDLTABLE.GGSCHEMA. PURGEDDLHISTORY.PURGEMARKERHIS ...