首页 > ACM题库 > HDU-杭电 > hdu 2602 Bone Collector-背包问题-[解题报告]C++
2014
02-10

hdu 2602 Bone Collector-背包问题-[解题报告]C++

Bone Collector

问题描述 :

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 ?

输入:

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.

输出:

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.

样例输入:

1
5 10
1 2 3 4 5
5 4 3 2 1

样例输出:

14

地址:http://acm.hdu.edu.cn/showproblem.php?pid=2602

思路:dp问题之01背包

代码如下:一维的,我习惯的作风。。。

#include<stdio.h>
 #include<stdlib.h>
 #include<string.h>
 #define N 1001
 int dp[N];
 int c[N],w[N];
 int max(int x,int y)
 {
     return x>y?x:y;
 }
 int main()
 {
     int t,n,v,i,j;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d%d",&n,&v);
         for(i=0;i<n;i++)
           scanf("%d",&c[i]);
         for(i=0;i<n;i++)
           scanf("%d",&w[i]);
         memset(dp,0,sizeof(dp));
         for(i=0;i<n;i++)
         {
             for(j=v;j>=w[i];j--)
               dp[j]=max(dp[j],dp[j-w[i]]+c[i]);
         }
         printf("%d\n",dp[v]);
     }
     system("pause");
     return 0;
 }

 

 

代码如下:二维的。。。不太习惯。。。不建议用

//二维数组实现01背包
 #include <iostream>
 #include <algorithm>
 using namespace std;
 int dp[1001][1001],vol[1001],val[1001];
 int main()
 {
     int t,n,i,j,v;
     cin>>t;
     while(t--)
     {
         cin>>n>>v;
         for(i=1;i<=n;i++)
             scanf("%d",&val[i]);
         for(i=1;i<=n;i++)
             scanf("%d",&vol[i]);
         memset(dp,0,sizeof(dp)); //初始化
         for(i=1;i<=n;i++)
             for(j=0;j<=v;j++) //注意要从0开始,这个题测试数据有点变态,有的骨头有价值,但占的空间是0
             {
                 if(vol[i]<=j && dp[i-1][j]<dp[i-1][j-vol[i]]+val[i])
                     dp[i][j]=dp[i-1][j-vol[i]]+val[i];
                 else
                     dp[i][j]=dp[i-1][j];
             }
         cout<<dp[n][v]<<endl;
     }
     return 0;
 }

 

 

解题转自:http://www.cnblogs.com/mycapple/archive/2012/08/22/2650279.html


  1. 换句话说,A[k/2-1]不可能大于两数组合并之后的第k小值,所以我们可以将其抛弃。
    应该是,不可能小于合并后的第K小值吧

  2. 嗯 分析得很到位,确实用模板编程能让面试官对你的印象更好。在设置辅助栈的时候可以这样:push时,比较要push的elem和辅助栈的栈顶,elem<=min.top(),则min.push(elem).否则只要push(elem)就好。在pop的时候,比较stack.top()与min.top(),if(stack.top()<=min.top()),则{stack.pop();min.pop();},否则{stack.pop();}.

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