【HDU】-2602-Bone Collector(01背包)

    xiaoxiao2025-08-01  5

    Bone Collector

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 51799    Accepted Submission(s): 21815 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   题解:01背包的基础题,详解见点击打开解析

    先来分析01背包:

    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)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。

    这是最基础的背包问题,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]=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。(时间复杂度不可小了,但是空间复杂度还可以减小)

    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!

     

    1 2 3 for  i=1..N     for  v=V..0          f[v]=max{f[v],f[v-c[i]]+w[i]};

     

     

     

    分析上面的代码:当内循环是逆序时,就可以保证后一个f[v]和f[v-c[i]]+w[i]是前一状态的!(不懂自己调试理解)

    AC代码:

    #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> using namespace std; #define CLR(a,b) memset(a,b,sizeof(a)) int main() { int u; int va[1010],v[1010]; int sum[1010]; scanf("%d",&u); while(u--) { int n,V; CLR(sum,0); scanf("%d %d",&n,&V); for(int i=1;i<=n;i++) scanf("%d",&va[i]); for(int i=1;i<=n;i++) scanf("%d",&v[i]); for(int i=1;i<=n;i++) { for(int j=V;j>=v[i];j--) sum[j]=max(sum[j],sum[j-v[i]]+va[i]);<span style="white-space:pre"> </span>//注意这里是max(sum[j],...) } printf("%d\n",sum[V]); } return 0; }

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