hdoj 2602 Bone Collector【0-1背包】【dp思想】

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 40247    Accepted Submission(s): 16722

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：给一定数量的骨头的价格和体积，装进背包，求所能装下的骨头的最大价值

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct node{int value,vol;}; node p[1010];int bag[1100];int main(){int i,j,k,t,n,v;scanf("%d",&t);//总情况数 while(t--){memset(bag, 0, sizeof(bag));//全初始化为 0 scanf("%d%d",&n,&v);//n数量,v体积 for(i = 1; i <= n; i++)scanf("%d",&p[i].value);//输入每个骨头的价值 for(i = 1; i <= n; i++)scanf("%d",&p[i].vol);//输入每个骨头所占体积 for(i = 1; i <= n ;i++)//数量变化 {for( j = v; j >= p[i].vol;j--)// 体积控制,背包如果可以装下骨头就继续 {//动态规划过程 bag[j] = max(bag[j], bag[j-p[i].vol]+p[i].value);//当前与下一个取价值最大的 }}printf("%d\n",bag[v]);//输出装的最多价值 }return 0;}