HDU2602_01背包问题

01背包（ZeroOnePack）: 

       有N件物品和一个容量为V的背包。（ 每种物品均只有一件 ，可以选择放或不放） 第i件物品的费用是c[i]，价值是w[i]。求解将哪些物品装入背包可使价值总和最大。 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]=max{f[i-1][v],f[i-1][v-c[i]]+w[i]}

把这个过程理解下：在前i件物品放进容量v的背包时，

它有两种情况：

第一种是第i件不放进去，这时所得价值为:f[i-1][v]

第二种是第i件放进去，这时所得价值为：f[i-1][v-c[i]]+w[i]

（第二种是什么意思？就是如果第i件放进去，那么在容量v-c[i]里就要放进前i-1件物品）

最后比较第一种与第二种所得价值的大小，哪种相对大，f[i][v]的值就是哪种。（这是基础，要理解！）

这里是用二位数组存储的，可以把空间优化，用一位数组存储。

用f[0..v]表示，f[v]表示把前i件物品放入容量为v的背包里得到的价值。把i从1~n(n件)循环后，最后f[v]表示所求最大值。*这里f[v]就相当于二位数组的f[i][v]。那么，如何得到f[i-1][v]和f[i-1][v-c[i]]+w[i]？（重点！思考）

首先要知道，我们是通过i从1到n的循环来依次表示前i件物品存入的状态。即：for i=1..N

现在思考如何能在是f[v]表示当前状态是容量为v的背包所得价值，而又使f[v]和f[v-c[i]]+w[i]标签前一状态的价值？

逆序！

这就是关键！

经典例题：

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 47043    Accepted Submission(s): 19603

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 2 ³¹).

 

Sample Input

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

 

Sample Output

14

【代码如下】：

# include # include # include # define max(x,y) x>y?x:y; int v[1005];//价值 int w[1005];//重量 int dp[1005][1005]; int main() {     int T;     int N,V;     scanf("%d", &T);     while(T--)     {         scanf("%d%d",&N,&V);         memset(dp,0,sizeof(dp));         for(int i=1; i<=N; i++)             scanf("%d",&v[i]);         for(int i=1; i<=N; i++)             scanf("%d",&w[i]);         for(int i=1; i<=N; i++)             for(int j=V; j>=w[i]; j--) //放入背包             {                 if(dp[i-])                 {                     dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+v[i]);                 }             }         printf("%d\n",dp[N][V]);     }     return 0; }