hdu 2602 Bone Collector（01背包）
Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 42179 Accepted Submission(s):
17543
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 ?
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.
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
1
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14
Author
Teddy
Source
Recommend
很久之前就写过这道题，很久之前也接触过背包，写过很多01背包的题，但是就是记不住！貌似对所有的dp我都不能灵活的运用，因此感到非常懊恼。于是我决定把所以的背包题重新写一次。
最简单的背包题，模板01背包。
题意：第一行输入几组数据，第二行第一个数字代表物体个数，第二个数代表总体积。需要注意的是，第三排输入的是物品的价值，第四排的物品的体积。在不可以拆分物体的前提下，已知背包的总体积，最大能获取的价值是多少。
附上代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 int max(int a,int b) 6 { 7 return a>b?a:b; 8 } 9 int main() 10 { 11 int T,n,m,i,j; 12 int a[1005],b[1005]; 13 scanf("%d",&T); 14 while(T--) 15 { 16 scanf("%d%d",&n,&m); 17 for(i=0; i<n; i++) //输入价值 18 scanf("%d",&a[i]); 19 for(i=0; i<n; i++) //输入体积 20 scanf("%d",&b[i]); 21 int dp[1005]; //dp数组始终记录当前体积的最大价值 22 memset(dp,0,sizeof(dp)); //dp全部初始化为0 23 for(i=0; i<n; i++) //从第一个开始循环 24 for(j=m; j>=b[i]; j--) 25 dp[j]=max(dp[j],dp[j-b[i]]+a[i]); //比较放入i物体后的价值与不放之前的价值，记录大的值 26 printf("%d\n",dp[m]); //输入总体积的最大价值 27 } 28 return 0; 29 }
