ACM 杭电hdu 2602 Bone Collector（01背包）

                                             Bone Collector

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

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

题意：   一个叫做Bone Collector的男的有一个包，往包里放东西，使得其价值最大。

输入：注意是先输入的是价值，后是体积。

现在开始搞背包问题，网上看了些资料。

01 背包问题特点是：每种物品仅有一件，可以选择放或不放。用子问题定义状态：即F[i;v] 表示前i 件物品

恰放入一个容量为v的背包可以获得的最大价值。则其状态转移方程便是：
F[i;v] = maxfF[i-1;v];F[i-1;v-Ci] + Wig
这个方程非常重要，基本上所有跟背包相关的问题的方程都是由它衍生出来的。所以有必要将它详细解释一下：

“将前i 件物品放入容量为v的背包中”这个子问题，若只考虑第i 件物品的策略（放或不放），那么就可以转

化为一个只和前i-1件物品相关的问题。如果不放第i 件物品，那么问题就转化为“前i-1件物品放入容量为v的

背包中”，价值为F[i-1; v]；如果放第i 件物品，那么问题就转化为“前i-1件物品放入剩下的容量为v-Ci 的

背包中”，此时能获得的最大价值就是F[i-1;v-Ci] 再加上通过放入第i 件物品获得的价值Wi。

为了方便理解这里还找了一个图。

刚刚接触这个问题不免还有有点儿不熟悉。

F[i-1,v]倒是好理解. F[i-1,v-Ci]+Wi 稍微难得看懂点。其实如果丢掉后面的 Wi，F[i-1,v-Ci]可以按照

F[i-1,v],一样理解。下面就贴代码了。。。

 

#include<stdio.h>#include<string.h>#define M 1009typedef struct pack{int cost;int val;}PACK;int f[M][M];int main(){int cas,n,v,i,j;PACK a[M];scanf("%d",&cas);while(cas--){scanf("%d%d",&n,&v);memset(f,0,sizeof(f));for(i=1;i<=n;i++)scanf("%d",&a[i].val);for(i=1;i<=n;i++)scanf("%d",&a[i].cost);for(i=1;i<=n;i++)for(j=0;j<=v;j++)if(j-a[i].cost>=0&&f[i-1][j]<f[i-1][j-a[i].cost]+a[i].val)f[i][j]=f[i-1][j-a[i].cost]+a[i].val;elsef[i][j]=f[i-1][j];printf("%d\n",f[n][v]);}return 0;} 

 

 

注意：两个for 循环，循环条件最好不要变，对于二维的。因为循环是一层一层开始更新的；

而一维的话是始终对同一层进行修改！

一维的代码有机会再补上。

 

 

#include<stdio.h>#include<string.h>#define M 1009typedef struct pack{int cost;int val;}PACK;int main(){int cas,n,v,i,j;int f[M];PACK a[M];scanf("%d",&cas);while(cas--){scanf("%d%d",&n,&v);memset(f,0,sizeof(f));for(i=1;i<=n;i++)scanf("%d",&a[i].val);for(i=1;i<=n;i++)scanf("%d",&a[i].cost);for(i=1;i<=n;i++)for(j=v;j>=a[i].cost;j--)if(f[j]<f[j-a[i].cost]+a[i].val)f[j]=f[j-a[i].cost]+a[i].val;printf("%d\n",f[v]);}return 0;}